Answer
Solution set: $\{ \frac{-2+\sqrt 10}{2}, \frac{-2-\sqrt 10}{2} \}$
Work Step by Step
$2x^{2} = 3-4x$
$2x^{2}+4x-3=0$
Using quadratic formula,
$x= \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
Substituting $a=2, b=4, c=-3$
$x= \frac{-4±\sqrt ((4)^{2}-4(2)(-3))}{2(2)}$
$x= \frac{-4±\sqrt (16+24)}{4}$
$x= \frac{-4±\sqrt 40}{4}$
$x= \frac{-4±\sqrt 4\times10 }{4}$
$x= \frac{-4±2\sqrt 10}{4}$
$x= \frac{2(-2±\sqrt 10)}{4}$
$x= \frac{(-2±\sqrt 10)}{2}$
$x= \frac{(-2+\sqrt 10)}{2}$ or $x= \frac{(-2-\sqrt 10)}{2}$
