# How do you find the roots, real and imaginary, of y=-x^2 +32x -16 using the quadratic formula?

Oct 7, 2017

$x = 16 + \sqrt{240} = 31.49$ OR $x = 16 - \sqrt{240} = .51$

There are no imaginary roots.

#### Explanation:

For $y = a {x}^{2} + b x + c$

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

In your problem, $a = - 1 , b = 32 , c = - 16$

So, $x = \frac{- 32 \pm \sqrt{{32}^{2} - 4 \left(- 1\right) \left(- 16\right)}}{2 \left(- 1\right)}$

Simplifying:

$x = \frac{- 32 \pm \sqrt{1024 - 64}}{- 2}$

$x = \frac{- 32 \pm \sqrt{960}}{- 2}$

$x = \frac{- 32}{-} 2 \pm \frac{\sqrt{4 \cdot 240}}{- 2}$

$x = 16 \pm \frac{2 \sqrt{240}}{- 2}$

So $x = 16 + \sqrt{240} = 31.49$ OR $x = 16 - \sqrt{240} = .51$

There are no imaginary roots.

Oct 7, 2017

Roots are $16 - 4 \sqrt{15}$ and $16 + 4 \sqrt{15}$

#### Explanation:

According to quadratic formula, the roots of $y = a {x}^{2} + b x + c$ are

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Hence roots of $y = - {x}^{2} + 32 x - 16$ are

$\frac{- 32 \pm \sqrt{{32}^{2} - 4 \times \left(- 1\right) \left(- 16\right)}}{2 \times \left(- 1\right)}$

= $\frac{- 32 \pm \sqrt{1024 - 64}}{- 2}$

= $16 \pm \frac{\sqrt{960}}{2}$

= $16 \pm \frac{8 \sqrt{15}}{2}$

= $16 \pm 4 \sqrt{15}$