(2x)^2+4x=120

Solution for (2x)^2+4x=120 equation:

(2x)^2+4x=120

We move all terms to the left:

(2x)^2+4x-(120)=0

a = 2; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·2·(-120)
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{61}}{2*2}=\frac{-4-4\sqrt{61}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{61}}{2*2}=\frac{-4+4\sqrt{61}}{4}
$