# What are the zeros -2x^2-15x+y+22=0?

May 21, 2018

$x = \frac{- 15 + \sqrt{401}}{4}$, $\frac{- 15 - \sqrt{401}}{4}$

#### Explanation:

Given:

$- 2 {x}^{2} - 15 x + y + 22 = 0$

Subtract $y$ from both sides.

$- 2 {x}^{2} - 15 x + 22 = - y$

Multiply both sides by $- 1$. This will reverse the signs.

$2 {x}^{2} + 15 x - 22 = y$

Switch sides.

$y = 2 {x}^{2} + 15 x - 22$

This is a quadratic equation in standard form:

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

where:

$a = 2$, $b = 15$, $c = - 22$

The roots are the x-intercepts, which are the values for $x$ when $y = 0$.

Substitute $0$ for $y$.

$0 = 2 {x}^{2} + 15 x - 22$

Solve for $x$ using the quadratic formula:

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

Plug the known values into the equation.

$x = \frac{- 15 \pm \sqrt{{15}^{2} - 4 \cdot 2 \cdot - 22}}{2 \cdot 2}$

$x = \frac{- 15 \pm \sqrt{401}}{4}$ $\leftarrow$ $401$ is a prime number

Roots

$x = \frac{- 15 + \sqrt{401}}{4}$, $\frac{- 15 - \sqrt{401}}{4}$

Approximate roots

$x \approx 2.56 ,$ $- 8.756$

graph{y=2x^2+15x-22 [-11.09, 11.41, -8.775, 2.475]}