By using quadratic formula:
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\quad a=1,\:b=20,\:c=-8[/tex]
[tex]x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\cdot \:1\left(-8\right)}}{2\cdot \:1}[/tex]
[tex]x=\frac{-20+\sqrt{20^2-4\cdot \:1\left(-8\right)}}{2\cdot \:1}:\quad 2\left(3\sqrt{3}-5\right)[/tex]
[tex]x=\frac{-20-\sqrt{20^2-4\cdot \:1\left(-8\right)}}{2\cdot \:1}:\quad -2\left(5+3\sqrt{3}\right)[/tex]
So the solutions are:
[tex]\boxed{x=2\left(3\sqrt{3}-5\right),\:x=-2\left(5+3\sqrt{3}\right)}[/tex]
