## College Algebra (10th Edition)

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