Answer
$x = \frac{-1 \frac{+}{} \sqrt {33}}{4}$
Work Step by Step
$$2x^{2} = 4 - x$$
$2x^{2} + x - 4 = 0$; where we can use the quadratic formula ($x = \frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$) to solve for $x$:
$x = \frac{-1 \frac{+}{} \sqrt {1^{2} - 4(2)(-4)}}{2(2)} = \frac{-1 \frac{+}{} \sqrt {1 + 32}}{4} = \frac{-1 \frac{+}{} \sqrt {33}}{4}$
