125=4x^2

Solution for 125=4x^2 equation:

125=4x^2

We move all terms to the left:

125-(4x^2)=0

a = -4; b = 0; c = +125;
Δ = b2-4ac
Δ = 02-4·(-4)·125
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*-4}=\frac{0-20\sqrt{5}}{-8}
=-\frac{20\sqrt{5}}{-8}
=-\frac{5\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*-4}=\frac{0+20\sqrt{5}}{-8}
=\frac{20\sqrt{5}}{-8}
=\frac{5\sqrt{5}}{-2}
$