# How do you solve x^2 + x = 3/4?

$\implies 4 {x}^{2} + 4 x - 3 = 0$
$\implies 4 {x}^{2} + 6 x - 2 x - 3 = 0$
$\implies 2 x \left(2 x + 3\right) - 1 \left(2 x + 3\right) = 0$
$\implies \left(2 x + 3\right) \left(2 x - 1\right) = 0$
$\therefore x = - \frac{3}{2} \mathmr{and} x = \frac{1}{2}$