Question

Ac method for x^2+20x-8=0

Answer 1

Explanation:

By using quadratic formula:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\quad a=1,\:b=20,\:c=-8$

$x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\cdot \:1\left(-8\right)}}{2\cdot \:1}$

$x=\frac{-20+\sqrt{20^2-4\cdot \:1\left(-8\right)}}{2\cdot \:1}:\quad 2\left(3\sqrt{3}-5\right)$

$x=\frac{-20-\sqrt{20^2-4\cdot \:1\left(-8\right)}}{2\cdot \:1}:\quad -2\left(5+3\sqrt{3}\right)$

So the solutions are:

$\boxed{x=2\left(3\sqrt{3}-5\right),\:x=-2\left(5+3\sqrt{3}\right)}$