GED Free
Español Get the app
Menu

Solving Quadratic Equations

Solve quadratic equations by square roots, factoring and the quadratic formula, and tell how many solutions there are.

Practice this skill 30 min Solve quadratic equations Leer en español
Key idea

What you'll learn

A quadratic equation has a squared variable, like x^2 - 2x - 3 = 0. In this lesson you will learn three ways to solve one:

  1. Square roots — for equations like x^2 = 49 or (x - 3)^2 = 16.
  2. Factoring — for equations like x^2 + 2x - 15 = 0 that split into two easy factors.
  3. The quadratic formula — works for every quadratic, and it is on the GED formula sheet.

You will also learn how to tell, before solving, whether an equation has two, one, or no real solutions.

Why it matters on the GED: quadratic equations appear in the algebra part of the test as "find the solutions," "select all solutions," "which value is a solution," and as the final step of area and motion word problems (the next lesson). Graphs of parabolas also show the solutions as x-intercepts.

Real life: the height of a thrown ball, the area of a garden, and business profit are often quadratic. Solving the equation tells you when the ball lands or what size gives the area you need.

Words to know

  • Quadratic equation — an equation whose highest power of the variable is 2. Example: 3x^2 - 12 = 63.
  • Standard form — ax^2 + bx + c = 0, with everything on one side and 0 on the other. Example: 2x^2 - 3x - 4 = 0 has a = 2, b = -3, c = -4.
  • Coefficient — the number in front of a variable. Example: in 5x^2, the coefficient is 5.
  • Solution (root, zero) — a value of x that makes the equation true. Example: x = 3 is a solution of x^2 = 9.
  • Square root — \sqrt{25} = 5 because 5^2 = 25. Every positive number has two square roots: +5 and -5, written \pm 5.
  • Factor — to rewrite as a product. Example: x^2 - 5x + 6 = (x - 2)(x - 3).
  • Zero product property — if A \cdot B = 0, then A = 0 or B = 0.
  • Discriminant — the number b^2 - 4ac under the square root in the quadratic formula. It tells how many real solutions there are.
  • Parabola — the U-shaped graph of y = ax^2 + bx + c. Its x-intercepts are the solutions of ax^2 + bx + c = 0.

All Math words in the glossary

1. Solving with square roots

Use this method when there is an x^2 (or a squared group like (x - 3)^2) but no separate x term.

Steps: get the squared part alone, then take the square root of both sides, and remember \pm.

Example: x^2 = 49 → x = \pm\sqrt{49} = \pm 7. Both 7 and -7 work, because 7^2 = 49 and (-7)^2 = 49.

Example: 3x^2 - 12 = 63

  • Add 12: 3x^2 = 75
  • Divide by 3: x^2 = 25
  • Square root: x = \pm 5

Example (a squared group): (x - 3)^2 = 16

  • Square root: x - 3 = \pm 4
  • Two equations: x - 3 = 4 gives x = 7; x - 3 = -4 gives x = -1.
Graph of y = (x - 3) squared, a parabola with vertex (3, 0), and the horizontal line y = 16. They meet at x = -1 and x = 7, the two solutions of (x - 3) squared = 16, because x - 3 = 4 or x - 3 = -4.xy−3−2−1123456789481216200y = 16x = −1x = 7vertex (3, 0)y = (x − 3)²x − 3 = 4 or x − 3 = −4 → x = 7 or x = −1

Example (not a perfect square): x^2 = 20 → x = \pm\sqrt{20} \approx \pm 4.47.

Example (no real solution): x^2 = -9. No real number squared is negative, so there is no real solution.

In a real-world problem where x is a length or time, keep only the positive solution.

2. Solving by factoring

Step 1 — Standard form. Move every term to one side so the equation equals 0. Example: x^2 + 8x = 9 becomes x^2 + 8x - 9 = 0.

Step 2 — Factor. For x^2 + bx + c (when a = 1), find two numbers that multiply to c and add to b.

EquationMultiply toAdd toNumbersFactored
x^2 - 5x + 6 = 06-5-2, -3(x - 2)(x - 3) = 0
x^2 + 2x - 15 = 0-1525, -3(x + 5)(x - 3) = 0
x^2 + 8x - 9 = 0-989, -1(x + 9)(x - 1) = 0
Box model for factoring x squared + 2x - 15. Columns are labelled x and +5, rows x and -3. The four cells are x squared, 5x, -3x and -15. The constants multiply to -15 and the middle terms add to 2x, so x squared + 2x - 15 = (x + 5)(x - 3), giving x = -5 or x = 3.Box (area) model: x² + 2x − 15 = (x + 5)(x − 3)x+5x−3x²5x−3x−15Corners: 5 × (−3) = −15. Middle: 5x + (−3x) = 2x.Solutions: x + 5 = 0 → x = −5; x − 3 = 0 → x = 3

Step 3 — Zero product property. Set each factor equal to 0 and solve. From (x + 5)(x - 3) = 0: x + 5 = 0 \Rightarrow x = -5, and x - 3 = 0 \Rightarrow x = 3.

Watch out: the signs flip. The factor (x + 5) gives the solution x = -5, not +5.

Special patterns:

  • Common factor: x^2 - 6x = 0 \Rightarrow x(x - 6) = 0 \Rightarrow x = 0 or x = 6. Don't divide by x — you'd lose the solution x = 0.
  • Difference of squares: x^2 - 81 = 0 \Rightarrow (x - 9)(x + 9) = 0 \Rightarrow x = \pm 9.

When a \ne 1: 5x^2 - 3x - 2 = 0 factors as (5x + 2)(x - 1) = 0, so 5x + 2 = 0 \Rightarrow x = -\tfrac{2}{5} and x = 1. Always check by expanding: (5x + 2)(x - 1) = 5x^2 - 5x + 2x - 2 = 5x^2 - 3x - 2 ✓. If factoring is hard, use the quadratic formula instead.

Using the graph. The solutions of x^2 - 2x - 3 = 0 are where the parabola y = x^2 - 2x - 3 crosses the x-axis: x = -1 and x = 3.

Coordinate plane from -5 to 5 on both axes showing the parabola y = x squared minus 2x minus 3. It opens upward, crosses the x-axis at x = -1 and x = 3, crosses the y-axis at y = -3, and has its lowest point (vertex) at (1, -4).-5-5-4-4-3-3-2-2-1-11122334455xyy = x² − 2x − 3

3. The quadratic formula and the discriminant

The quadratic formula solves any equation in standard form ax^2 + bx + c = 0:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Example: x^2 - 6x + 2 = 0, so a = 1, b = -6, c = 2.

  • -b = -(-6) = 6
  • b^2 - 4ac = (-6)^2 - 4(1)(2) = 36 - 8 = 28
  • x = \dfrac{6 \pm \sqrt{28}}{2}
  • \sqrt{28} \approx 5.2915, so x \approx \dfrac{6 + 5.2915}{2} \approx 5.65 or x \approx \dfrac{6 - 5.2915}{2} \approx 0.35.

Careful with signs: if b is negative, -b is positive. And b^2 is always positive or zero: (-6)^2 = 36, not -36. If c is negative, -4ac becomes plus: for 2x^2 - 3x - 4 = 0, b^2 - 4ac = 9 - 4(2)(-4) = 9 + 32 = 41.

The discriminant b^2 - 4ac tells you how many real solutions there are:

b^2 - 4acNumber of real solutionsGraph
positive2crosses the x-axis twice
zero1touches the x-axis once
negative0 (no real solutions)never reaches the x-axis
Three parabolas showing the discriminant. Left: y = x squared - 6x + 2, discriminant 28, crosses the x-axis twice near 0.35 and 5.65. Middle: y = 4x squared - 12x + 9, discriminant 0, touches the x-axis once at 1.5. Right: y = x squared + 4x + 7, discriminant -12, never reaches the x-axis, so no real solutions.xy246−8−4480≈0.35≈5.65x² − 6x + 2 = 0b² − 4ac = 28 > 02 solutionsxy−11234−8−44801.54x² − 12x + 9 = 0b² − 4ac = 01 solutionxy−5−4−3−2−11−8−4480x² + 4x + 7 = 0b² − 4ac = −12 < 0no real solutions

Example: x^2 + 4x + 7 = 0: 16 - 28 = -12 < 0 → no real solutions.

Example: 4x^2 - 12x + 9 = 0: 144 - 144 = 0 → exactly one solution, x = \tfrac{-b}{2a} = \tfrac{12}{8} = 1.5.

Which method? No x term → square roots. Easy numbers → try factoring. Anything else (or a decimal answer is asked for) → the quadratic formula.

Formula

The quadratic formula (on the formula sheet)

\text{If } ax^2 + bx + c = 0, \text{ then } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • a = coefficient of x^2 (cannot be 0)
  • b = coefficient of x (use 0 if there is no x term)
  • c = the constant term
  • \pm = do it once with + and once with - to get two solutions

The quadratic formula is on the GED formula sheet, so you don't need to memorize it — but you do need to put the equation in standard form (equal to 0) before reading off a, b and c. The discriminant idea (b^2 - 4ac positive / zero / negative) and the zero product property are not on the sheet.

Worked example

Worked example 1 (easy): square roots

Solve 2x^2 = 98.

  1. Divide both sides by 2: x^2 = 49.
  2. Take the square root of both sides: x = \pm\sqrt{49}.
  3. x = 7 or x = -7.
Graph of y = 2x squared and the line y = 98. They meet at x = -7 and x = 7, the two solutions of 2x squared = 98.xy−9−6−3369204060801001200y = 98x = −7x = 7y = 2x²2x² = 98 → x² = 49 → x = ±7

Answer: x = \pm 7.

Check: 2(7)^2 = 2(49) = 98 ✓ and 2(-7)^2 = 2(49) = 98 ✓

Worked example

Worked example 2: factoring

Solve x^2 - x - 12 = 0.

  1. Already in standard form. Find two numbers that multiply to -12 and add to -1: -4 and 3.
  2. Factor: (x - 4)(x + 3) = 0.
  3. Zero product property: x - 4 = 0 \Rightarrow x = 4; x + 3 = 0 \Rightarrow x = -3.
Graph of the parabola y = x squared - x - 12 = (x - 4)(x + 3). It crosses the x-axis at x = -3 and x = 4, the solutions of x squared - x - 12 = 0. Its vertex is at (0.5, -12.25).xy−5−4−3−2−1123456−12−8−4480x = −3x = 4y = x² − x − 12

Answer: x = 4 or x = -3.

Check: 4^2 - 4 - 12 = 16 - 16 = 0 ✓; (-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0 ✓

Worked example

Worked example 3: the quadratic formula

Solve 2x^2 + 5x - 3 = 0.

  1. a = 2, b = 5, c = -3.
  2. Discriminant: b^2 - 4ac = 5^2 - 4(2)(-3) = 25 + 24 = 49.
  3. x = \dfrac{-5 \pm \sqrt{49}}{2(2)} = \dfrac{-5 \pm 7}{4}.
  4. x = \dfrac{-5 + 7}{4} = \dfrac{2}{4} = \dfrac{1}{2} or x = \dfrac{-5 - 7}{4} = \dfrac{-12}{4} = -3.
Graph of the parabola y = 2x squared + 5x - 3. It crosses the x-axis at x = -3 and x = 1/2, the solutions found with the quadratic formula. Its vertex is at (-1.25, -6.125).xy−5−4−3−2−1123−8−4480x = −3x = ½y = 2x² + 5x − 3

Answer: x = \tfrac{1}{2} or x = -3.

Check: 2(\tfrac{1}{2})^2 + 5(\tfrac{1}{2}) - 3 = 0.5 + 2.5 - 3 = 0 ✓; 2(9) + 5(-3) - 3 = 18 - 15 - 3 = 0 ✓

Worked example

Worked example 4 (real world): a falling object

A tool is dropped from a roof. Its height, in feet, after t seconds is h = 144 - 16t^2. When does it hit the ground?

  1. On the ground, h = 0: 144 - 16t^2 = 0.
  2. Add 16t^2: 16t^2 = 144.
  3. Divide by 16: t^2 = 9.
  4. Square root: t = \pm 3. Time can't be negative, so t = 3.
Graph of the height h = 144 - 16t squared of a dropped tool, from 144 feet at t = 0 down to the ground (h = 0) at t = 3 seconds. The negative solution t = -3 is rejected.t (s)h (ft)0.511.522.533.540801201600start: 144 ftground at t = 3 sh = 144 − 16t²144 − 16t² = 0 → t² = 9 → t = 3 (t = −3 rejected)

Answer: 3 seconds.

Check: 144 - 16(3)^2 = 144 - 144 = 0 ✓

Worked example

Worked example 5 (GED-hard): work backwards

One solution of x^2 + kx - 12 = 0 is x = 3. Find k and the other solution.

  1. Substitute x = 3: 3^2 + 3k - 12 = 0, so 9 + 3k - 12 = 0.
  2. Simplify: 3k - 3 = 0 \Rightarrow k = 1.
  3. The equation is x^2 + x - 12 = 0. Factor: (x + 4)(x - 3) = 0.
  4. Solutions: x = 3 (given) and x = -4.
Graph of y = x squared + x - 12, the equation once k = 1 is found. It crosses the x-axis at the given solution x = 3 and the other solution x = -4. Product of the solutions: 3 times -4 = -12.xy−6−5−4−3−2−112345−12−8−4480x = −4 (other)x = 3 (given)y = x² + x − 12

Answer: k = 1; the other solution is x = -4.

Check: (-4)^2 + (-4) - 12 = 16 - 4 - 12 = 0 ✓. Shortcut: when a = 1, the two solutions multiply to c: 3 \times (-4) = -12 ✓

Calculator tip

Calculator tip: TI-30XS MultiView

  • Square roots: press 2nd then x² to get \sqrt{\ }. Example: 2nd x² 28 enter → 2\sqrt{7}; press toggle → 5.291502622.
  • Negative numbers: use the (−) key, not the minus key, for a negative value like b = -6.
  • Squaring: use the x² key. Put negatives in parentheses: ( (−) 6 ) x² → 36. Without parentheses, -6^2 is read as -36.
  • Whole formula: find the square root first (2nd x² 28 enter → 2\sqrt{7}, then toggle → 5.291502622), then compute each solution with parentheses around the top: ( 6 + 5.2915 ) ÷ 2 enter → about 5.65, and ( 6 − 5.2915 ) ÷ 2 enter → about 0.35.
  • Check a solution: substitute it back into the original equation; the result should be 0 (or very close because of rounding).

When NOT to use it: if the equation factors easily (like x^2 - x - 12 = 0), factoring by hand is faster. Use the calculator for the discriminant and for decimal answers.

Common mistake

Common mistakes

  • Forgetting the negative root. x^2 = 49 has two solutions, 7 and -7 — not just 7.
  • Signs flip when factoring. (x + 5)(x - 3) = 0 gives x = -5 and x = 3, not 5 and -3.
  • Not setting the equation to 0 first. For x^2 + 8x = 9 you can't factor x^2 + 8x and set it to 9. Rewrite as x^2 + 8x - 9 = 0. The same applies before using the formula: c is -9, not 9.
  • Sign errors in the formula. If b = -3, then -b = +3. And -4ac with a negative c becomes positive: -4(2)(-4) = +32.
  • Dividing by 2 instead of 2a. The whole numerator is divided by 2a. For 2x^2 - 3x - 4 = 0, the denominator is 2(2) = 4.
  • Forgetting to take the square root. (x - 3)^2 = 16 means x - 3 = \pm 4, not x - 3 = 16.
Test strategy

GED test strategy

  • Plug in the answer choices. A solution makes the equation equal 0. Substituting is often faster than solving, especially on "select all" questions.
  • Read what is asked: both solutions, the positive one, the smaller one, or the number of solutions? On word problems, discard negative lengths or times.
  • Use the discriminant for "how many solutions" questions — don't solve the whole thing.
  • Look at the formula sheet for the quadratic formula; copy it carefully and write a, b, c (with signs) before substituting.
  • Use the graph: if a parabola is shown, the solutions are the x-intercepts.
  • Round only at the end, and only to the place the question asks for.

Check yourself

Solve quadratic equationsNo calculator

What are the solutions of the equation x^2 = 64?

Answer choices
Hint

Which numbers, when multiplied by themselves, give 64? There is more than one.

Check yourself

Solve quadratic equationsFill in

Solve 2x^2 + 5 = 77.

What is the positive solution?

x =

Type your answer in the box. Fractions like 3/4 are fine.

Hint

Get x^2 alone on one side first (undo the addition, then the multiplication).

Summary

  • No x term: isolate the square, then take \pm\sqrt{\ }.
  • Factoring: set the equation to 0, factor, then set each factor to 0 (signs flip).
  • Quadratic formula (on the formula sheet): x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} — works every time.
  • Discriminant b^2 - 4ac: positive → 2 solutions, zero → 1, negative → none.
  • Always check a solution by substituting it back. The solutions are the x-intercepts of the parabola.
Practice this skill →