4x^2+40x=20

Solution for 4x^2+40x=20 equation:

4x^2+40x=20

We move all terms to the left:

4x^2+40x-(20)=0

a = 4; b = 40; c = -20;
Δ = b2-4ac
Δ = 402-4·4·(-20)
Δ = 1920
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:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{30}}{2*4}=\frac{-40-8\sqrt{30}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{30}}{2*4}=\frac{-40+8\sqrt{30}}{8}
$