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Learn to solve quadratic equations using factoring and the quadratic formula, with clear steps, common mistakes, and practice problems.
Quadratic equations appear in many algebra and function topics. This guide focuses on two reliable methods: factoring and the quadratic formula. You will see how to choose a method, carry out each step, and check your answers. The examples use simple numbers so you can follow the arithmetic. A convention for handwritten work is defined so that valid alternative layouts are accepted. All practice tasks are self-contained; any needed diagram is described in words.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Factoring works best when the quadratic expression factors neatly with integers. Start by writing the equation in standard form ax^2 + bx + c = 0. Then factor the left side into a product of two binomials. Use the zero product property: if a product equals zero, at least one factor must be zero. Set each factor equal to zero and solve for x.
For example, to solve x^2 - 5x + 6 = 0, find two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3, so the factored form is (x - 2)(x - 3) = 0. Setting each factor to zero gives x = 2 or x = 3. Always substitute back into the original equation to check.
The quadratic formula solves any quadratic equation in standard form: x = [-b ± sqrt(b^2 - 4ac)] / (2a). Identify a, b, and c carefully, including signs. Compute the discriminant b^2 - 4ac first. If it is negative, there are no real solutions. If it is zero, there is one repeated real solution. If it is positive, there are two distinct real solutions.
For instance, solve 2x^2 + 3x - 2 = 0. Here a = 2, b = 3, c = -2. The discriminant is 3^2 - 4(2)(-2) = 9 + 16 = 25. Then x = [-3 ± sqrt(25)] / (2*2) = (-3 ± 5)/4. This gives x = 2/4 = 0.5 and x = -8/4 = -2. Check both values in the original equation.
If the quadratic factors easily with integers, factoring is often faster. If factoring is difficult or impossible with integers, use the quadratic formula. Both methods require the equation in standard form. Checking is not optional: substitute each solution back into the original equation and verify that both sides are equal.
When writing by hand, use a clear convention: write the equation on one line, show factoring or substitution on the next lines, and box final answers. X-height is measured from baseline to midline; letter styles vary, so accept any legible handwriting that follows the stated layout. Numerical values in examples are hypothetical and chosen for easy arithmetic.
Solve x^2 - 4x - 5 = 0 by factoring, and verify the solutions.
The solutions are x = 5 and x = -1.
Why: Find two numbers that multiply to -5 and add to -4: -5 and 1. Factor as (x - 5)(x + 1) = 0. Set each factor to zero: x - 5 = 0 gives x = 5; x + 1 = 0 gives x = -1. Check: for x = 5, 25 - 20 - 5 = 0; for x = -1, 1 + 4 - 5 = 0.
Solve x^2 - 7x + 10 = 0 by factoring.
x = 5 or x = 2.
Why: Find two numbers that multiply to 10 and add to -7: -5 and -2. Factor as (x - 5)(x - 2) = 0. Set each factor to zero: x - 5 = 0 gives x = 5; x - 2 = 0 gives x = 2.
Check: Substitute x = 5: 25 - 35 + 10 = 0. Substitute x = 2: 4 - 14 + 10 = 0. Both checks pass.
Solve 2x^2 + 5x - 3 = 0 using the quadratic formula.
x = 0.5 or x = -3.
Why: Here a = 2, b = 5, c = -3. Discriminant = 5^2 - 4(2)(-3) = 25 + 24 = 49. Then x = [-5 ± sqrt(49)] / (2*2) = (-5 ± 7)/4. This gives x = 2/4 = 0.5 and x = -12/4 = -3.
Check: Substitute x = 0.5: 2(0.25) + 2.5 - 3 = 0.5 + 2.5 - 3 = 0. Substitute x = -3: 2(9) - 15 - 3 = 18 - 18 = 0. Both checks pass.
Solve x^2 + 6x + 9 = 0 by factoring.
x = -3 (repeated real solution).
Why: The expression factors as (x + 3)(x + 3) = 0, so x + 3 = 0 gives x = -3. Since the factor repeats, there is one repeated real solution.
Check: Substitute x = -3: 9 - 18 + 9 = 0. The check passes.
Solve 3x^2 - 4x + 1 = 0 using the quadratic formula.
x = 1 or x = 1/3.
Why: Here a = 3, b = -4, c = 1. Discriminant = (-4)^2 - 4(3)(1) = 16 - 12 = 4. Then x = [4 ± sqrt(4)] / (2*3) = (4 ± 2)/6. This gives x = 6/6 = 1 and x = 2/6 = 1/3.
Check: Substitute x = 1: 3 - 4 + 1 = 0. Substitute x = 1/3: 3(1/9) - 4(1/3) + 1 = 1/3 - 4/3 + 1 = -1 + 1 = 0. Both checks pass.
Yes, for any quadratic equation in standard form with real coefficients, the quadratic formula gives all real solutions, if any exist.
Try to find two integers that multiply to ac and add to b. If no such integers exist, factoring with integers is not possible, so use the formula.
Then the quadratic has one repeated real solution, x = -b/(2a).
Yes. Substituting each solution into the original equation confirms accuracy and catches sign or arithmetic errors.