# How do you find the zeros of  y = -2x^2 + 8x -2  using the quadratic formula?

Mar 15, 2018

$x = 2 \pm \sqrt{3}$

#### Explanation:

The Quadratic Formula states that if a quadratic is of the form $a {x}^{2} + b x + c$, it's zeroes are at

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

Our quadratic is in standard form, therefore we know

$a = - 2$
$b = 8$
$c = - 2$

Now, we can just plug in. We get:

(-8+-sqrt(8^2-4(-2)(-2)))/(2(-2)

Which simplifies to

$\frac{- 8 \pm \sqrt{64 - 16}}{-} 4$

Which can be further simplified to

$\frac{- 8 \pm \sqrt{48}}{-} 4$

At this point, we can factor $\sqrt{48}$ into $\textcolor{b l u e}{\sqrt{16} \cdot \sqrt{3}}$, because $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$. This simplifies to

$\frac{- 8 \pm \textcolor{b l u e}{4 \sqrt{3}}}{-} 4$

We can factor out a $4$ because all of the terms have a $4$ in common. We get:

$\frac{- 2 \pm \sqrt{3}}{-} 1$

Which can be simplified as

$2 \pm \sqrt{3}$