# How do you find the zeros, real and imaginary, of y=-x^2-2x+9 using the quadratic formula?

Apr 14, 2017

The Quadratic Formula is very useful in finding the roots of an equation. The formula is:
$\frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$
Where $a , b$ and $c$ come from ${a}^{2} x + b x + c$

So, in our equation ($y = - {x}^{2} - 2 x + 9$), $\textcolor{p u r p \le}{a} = \textcolor{p u r p \le}{- 1}$, $\textcolor{g r e e n}{b} = \textcolor{g r e e n}{- 2}$, and $\textcolor{red}{c} = \textcolor{red}{9}$

Thus, our equation is:
$\frac{- \left(\textcolor{g r e e n}{- 2}\right) \pm \sqrt{{\left(\textcolor{g r e e n}{- 2}\right)}^{2} - 4 \cdot \left(\textcolor{p u r p \le}{- 1}\right) \cdot \left(\textcolor{red}{9}\right)}}{2 \cdot \textcolor{p u r p \le}{- 1}}$
$\frac{2 \pm \sqrt{4 - - 36}}{- 2}$
$\frac{2 \pm \sqrt{40}}{-} 2$
$\frac{2 \pm 2 \sqrt{10}}{-} 2$
$\frac{\cancel{2} \left(1 \pm \sqrt{10}\right)}{\cancel{- 2}}$
$1 \pm \sqrt{10}$

Thus, $x = 1 \pm \sqrt{10}$, which is about $4.162$ and $2.162$. Those are the roots of $y = - {x}^{2} - 2 x + 9$)