# How do you find the zeros of  y = -2x^2 - 11x + 17  using the quadratic formula?

Mar 24, 2018

$x = - \frac{11}{4} + \frac{\sqrt{257}}{4}$
$x = - \frac{11}{4} - \frac{\sqrt{257}}{4}$

$x \approx 1.258$
$x \approx - 6.758$

#### Explanation:

$a {x}^{2} + b x + c$

Therefore in this equation:
$a = - 2$
$b = - 11$
$c = 17$

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

Plug in givens and solve:
$x = \frac{11 \pm \sqrt{{\left(- 11\right)}^{2} - 4 \left(- 2\right) \left(17\right)}}{2 \cdot - 2}$

$x = \frac{11 \pm \sqrt{121 + 136}}{- 4}$

$x = \frac{11 \pm \sqrt{257}}{- 4}$

$x = - \frac{11}{4} + \frac{\sqrt{257}}{4}$
$x = - \frac{11}{4} - \frac{\sqrt{257}}{4}$

$x \approx 1.258$
$x \approx - 6.758$

Graph to confirm:
graph{-2x^2-11x+17 [-10, 10, -5, 5]}

Mar 24, 2018

$x = - \frac{11 \pm \sqrt{257}}{4}$ or $x = - \frac{11 + \sqrt{257}}{4} , - \frac{11 - \sqrt{257}}{4}$

#### Explanation:

The quadratic formula is $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

This equation is written in standard form: $y = a {x}^{2} + b x + c$

To find the zeros using the quadratic formula, let's first establish what a, b, and c are.

$y = - 2 {x}^{2} - 11 x + 17 \rightarrow$ -2 is a, -11 is b, 17 is c

Plug these numbers into the formula:

$\frac{- \left(- 11\right) \pm \sqrt{{\left(- 11\right)}^{2} - 4 \cdot - 2 \cdot 17}}{2 \cdot - 2}$

$\frac{11 \pm \sqrt{121 - 4 \cdot - 2 \cdot 17}}{2 \cdot - 2}$

$\frac{11 \pm \sqrt{121 - \left(- 136\right)}}{-} 4$

$\frac{11 \pm \sqrt{257}}{-} 4$

$- \frac{11 \pm \sqrt{257}}{4} \rightarrow$ This is your final answer because the square root of 257 cannot be reduced

The answer to the equation is $x = - \frac{11 \pm \sqrt{257}}{4}$ or $x = - \frac{11 + \sqrt{257}}{4} , - \frac{11 - \sqrt{257}}{4}$. These are the zeros (when plugged in for x, they make the function zero).