Answer
$\displaystyle \frac{1\pm\sqrt{13}}{4}$
Work Step by Step
We solve:
$2x+3=4x^{2}$
$2x+3-4x^{2}=0$
$4x^{2}-2x-3=0$
We solve using the quadratic formula ($a=4,\ b=-2,\ c=-3$):
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4*4*-3}}{2(4)}
=\frac{2\pm\sqrt{52}}{8}=\frac{2\pm 2\sqrt{13}}{8}=\frac{1\pm\sqrt{13}}{4}$
