20x2+20x-5=0

Solution for 20x2+20x-5=0 equation:

20x^2+20x-5=0

a = 20; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·20·(-5)
Δ = 800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*20}=\frac{-20-20\sqrt{2}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*20}=\frac{-20+20\sqrt{2}}{40}
$