4x(3x+40)=20

Solution for 4x(3x+40)=20 equation:

4x(3x+40)=20

We move all terms to the left:

4x(3x+40)-(20)=0

We multiply parentheses

12x^2+160x-20=0

a = 12; b = 160; c = -20;
Δ = b2-4ac
Δ = 1602-4·12·(-20)
Δ = 26560
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:

$\sqrt{\Delta}=\sqrt{26560}=\sqrt{64*415}=\sqrt{64}*\sqrt{415}=8\sqrt{415}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{415}}{2*12}=\frac{-160-8\sqrt{415}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{415}}{2*12}=\frac{-160+8\sqrt{415}}{24}
$