# What are the x intercepts for f(x) = -2x^2 + 4x + 3?

May 22, 2015

f(x) = -2x^2 + 4x + 3 = 0

$D = {d}^{2} = {b}^{2} - 4 a c = 16 + 24 = 40 = 4 \left(10\right) \to d = \pm 2 \sqrt{10}$

Two x-intercepts: $x = - \frac{b}{2} a \pm \frac{d}{2} a$

$x = 1 + \frac{2 \sqrt{10}}{-} 4 = 1 - \frac{\sqrt{10}}{2}$

$x = 1 + \frac{\sqrt{10}}{2}$