How to use the quadratic formula

The quadratic formula finds the roots of a quadratic equation by evaluating x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. It applies to any polynomial equation of degree two that is expressed in the standard form ax2+bx+c=0ax^2 + bx + c = 0.

The setup

Rearrange the equation into standard form ax2+bx+c=0ax^2 + bx + c = 0. Identify the constant coefficients aa, bb, and cc. Ensure that aeq0a eq 0, otherwise the equation is linear, not quadratic.

The steps

  1. Calculate the discriminant Δ=b24ac\Delta = b^2 - 4ac. 2. Substitute aa, bb, and Δ\Delta into the quadratic formula x=b±Δ2ax = \frac{-b \pm \sqrt{\Delta}}{2a}. 3. Evaluate the positive and negative cases to find the two roots.

Checking the result

Substitute the calculated values of xx back into the original equation ax2+bx+c=0ax^2 + bx + c = 0. The left side of the equation must evaluate exactly to zero.

Common errors

Failing to put the equation in standard form before identifying coefficients is a primary source of error. Other common mistakes include dropping negative signs when squaring bb or evaluating 4ac-4ac, and incorrectly canceling terms across the numerator and denominator.

Worked example

Find the roots of 2x25x=32x^2 - 5x = 3.

Standard form: 2x25x3=02x^2 - 5x - 3 = 0. Identify variables: a=2a = 2, b=5b = -5, c=3c = -3. Calculate discriminant: Δ=(5)24(2)(3)=25+24=49\Delta = (-5)^2 - 4(2)(-3) = 25 + 24 = 49. Apply formula: x=(5)±492(2)x = \frac{-(-5) \pm \sqrt{49}}{2(2)}. Simplify numerator and denominator: x=5±74x = \frac{5 \pm 7}{4}. Calculate first root: x1=124=3x_1 = \frac{12}{4} = 3. Calculate second root: x2=24=0.5x_2 = \frac{-2}{4} = -0.5.

FAQ

Run your own problem

References: OpenStax College Algebra, Chapter 2.5 · Khan Academy, Quadratic equations

See also