4x^2+20x+5=0

Solution for 4x^2+20x+5=0 equation:

4x^2+20x+5=0

a = 4; b = 20; c = +5;
Δ = b2-4ac
Δ = 202-4·4·5
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{5}}{2*4}=\frac{-20-8\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{5}}{2*4}=\frac{-20+8\sqrt{5}}{8}
$