4x^2+64x+20=0

Solution for 4x^2+64x+20=0 equation:

4x^2+64x+20=0

a = 4; b = 64; c = +20;
Δ = b2-4ac
Δ = 642-4·4·20
Δ = 3776
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:

$\sqrt{\Delta}=\sqrt{3776}=\sqrt{64*59}=\sqrt{64}*\sqrt{59}=8\sqrt{59}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{59}}{2*4}=\frac{-64-8\sqrt{59}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{59}}{2*4}=\frac{-64+8\sqrt{59}}{8}
$