13=4x2

Solution for 13=4x2 equation:

13=4x^2

We move all terms to the left:

13-(4x^2)=0

a = -4; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-4)·13
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*-4}=\frac{0-4\sqrt{13}}{-8}
=-\frac{4\sqrt{13}}{-8}
=-\frac{\sqrt{13}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*-4}=\frac{0+4\sqrt{13}}{-8}
=\frac{4\sqrt{13}}{-8}
=\frac{\sqrt{13}}{-2}
$