# How do you find the x intercepts for y = -2x^2 + 5x - 1?

Jun 30, 2015

Solve $y = - 2 {x}^{2} + 5 x - 1 = 0$

#### Explanation:

$D = {d}^{2} = 25 - 8 = 17 \to d = \pm \sqrt{17}$

$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4}$

Jul 1, 2015

$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4}$

#### Explanation:

This answer is the same as given by Nghi N. I am simply providing a more detailed explanation.

Since the x-intercepts occur when $y = 0$ we solve the given equation with $0$ in place of $y$ to find the x-intercepts.

−2x^2+5x−1=0

Any parabolic equation of the form $a {x}^{2} + b x + c = 0$
can be solved using the quadratic formula:
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Applying the values from our example:
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(- 2\right) \left(- 1\right)}}{2 \left(- 2\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 5 \pm \sqrt{25 - 8}}{- 4}$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4}$