Answer
$$x=\frac{1+\sqrt{13}}{4},\:x=\frac{1-\sqrt{13}}{4}$$
Work Step by Step
Solving by plugging in the given value and then using the quadratic formula, we find:
$$ 4x^2-2x-3=0\\ x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:4\left(-3\right)}}{2\cdot \:4}\\ x=\frac{1+\sqrt{13}}{4},\:x=\frac{1-\sqrt{13}}{4}$$
