# How do you solve y = x^2 - 16x + 58 using the quadratic formula?

Mar 6, 2018

$x = 8 \pm \sqrt{6}$

#### Explanation:

For every $A {x}^{2} + B x + C = 0$,

$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$

For the equation ${x}^{2} - 16 x + 58 = 0$

$x = \frac{- \left(- 16\right) \pm \sqrt{{\left(- 16\right)}^{2} - 4 \left(1\right) \left(58\right)}}{2 \left(1\right)}$

$x = \frac{16 \pm \sqrt{256 - 232}}{2}$

$x = \frac{16 \pm \sqrt{24}}{2}$

$x = \frac{16 \pm 2 \sqrt{6}}{2}$

$x = 8 \pm \sqrt{6}$

Note: the sign $\pm$ means that both addition and subtraction gives answers, in this case, the two roots are $8 + \sqrt{6}$ and $8 - \sqrt{6}$.

Mar 6, 2018

$8 \pm \sqrt{6}$

which would mean $8 + \sqrt{6}$ and $8 - \sqrt{6}$ are your two solutions.

#### Explanation:

Alright, so our quadratic formula is

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

So our $a$, $b$, and $c$ values are all the leading coefficients, so $\left(1 , - 16 , 58\right)$

Next step: insert the numbers into the formula (double negative makes the $16$ positive)

${x}_{1 , 2} = \frac{- \left(- 16\right) \pm \sqrt{- {16}^{2} - 4 \left(1\right) \left(58\right)}}{2 \left(1\right)}$

${x}_{1 , 2} = \frac{8 \pm \sqrt{24}}{2}$

Now we can simplify our square root by doing

$\sqrt{24} = 2 \sqrt{6}$

The two's will now cancel out on the top and bottom and we will be left with

${x}_{1 , 2} = 8 \pm \sqrt{6}$