Checkpoint: The Quadratic Formula Solution

Solve.

$x^{2} + 29 = 4 x$

$x^{2} - 4 x + 29 = 0 a = 1, b = - 4, c = 29 x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (29)}}{2 (1)} x = \frac{4 \pm \sqrt{16 - 116}}{2} = \frac{4 \pm \sqrt{- 100}}{2} = \frac{4 \pm i \sqrt{100}}{2} x = \frac{4 \pm 10 i}{2} = \frac{2 (2 \pm 5 i)}{2} x = 2 \pm 5 i$

Note

Q: Why is it helpful to name a, b, and c before substituting them into the quadratic formula?

A: To make sure that the numbers being substituted are correctly identified.