(2x)^2=240

Solution for (2x)^2=240 equation:

(2x)^2=240

We move all terms to the left:

(2x)^2-(240)=0

a = 2; b = 0; c = -240;
Δ = b2-4ac
Δ = 02-4·2·(-240)
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{-(0)-8\sqrt{30}}{2*2}=\frac{0-8\sqrt{30}}{4}
=-\frac{8\sqrt{30}}{4}
=-2\sqrt{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{30}}{2*2}=\frac{0+8\sqrt{30}}{4}
=\frac{8\sqrt{30}}{4}
=2\sqrt{30}
$