Answer
Solution set: $\{-1-\sqrt{3}, -1+\sqrt{3}\}$
Work Step by Step
Multiply both sides with $-2$
(2 is the common denominator,
and we like the leading term to be positive)
$x^{2}+2x-2=0$
Quadratic formula, a=$1$, b=$2$, c=$-2$:
$x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Checking $\Delta=b^{2}-4ac=2^{2}-4(1)(-2)=4+8=12$
$\Delta$ is positive, so there will be 2 real solutions.
$x=\displaystyle \frac{-(2)\pm\sqrt{12}}{2(1)}=\frac{-2\pm\sqrt{4\cdot 3}}{2}$
$=\displaystyle \frac{-2\pm 2\sqrt{3}}{2}=\frac{2(-1\pm\sqrt{3})}{2}=-1\pm\sqrt{3}$
$x=-1-\sqrt{3}$
or
$x=-1+\sqrt{3}$
Solution set: $\{-1-\sqrt{3}, -1+\sqrt{3}\}$
