Answer
$\left\{\dfrac{-1- \sqrt{5}}{4},\dfrac{-1+ \sqrt{5}}{4}\right\}$
Work Step by Step
Add $2x-1$ to both sides.
$4x^2+2x-1=1-2x+2x-1$
Simplify.
$4x^2+2x-1=0$
The equation is in standard form $ax^2+bx+c=0$.
We have $a=4,b=2$ and $c=-1$.
The discriminant is
$=b^2-4ac$
$=(2)^2-4(4)(-1)$
$=4+16$
$=20$
Since $20<0$. There are two real solutions.
The quadratic formula is
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
Substitute the values.of $a, b, c, $ and the discriminant to obtain:
$x=\dfrac{-(2)\pm \sqrt{(20)}}{2(4)}$
Simplify.
$x=\dfrac{-2\pm 2\sqrt{5}}{8}$
Factor out $2$.
$x=\dfrac{2(-1\pm \sqrt{5})}{8}$
Simplify.
$x=\dfrac{-1\pm \sqrt{5}}{4}$
Hence, the solution set is $\left\{\dfrac{-1- \sqrt{5}}{4},\dfrac{-1+ \sqrt{5}}{4}\right\}$.
