13=5s2

Solution for 13=5s2 equation:

13=5s^2

We move all terms to the left:

13-(5s^2)=0

a = -5; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-5)·13
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{65}}{2*-5}=\frac{0-2\sqrt{65}}{-10}
=-\frac{2\sqrt{65}}{-10}
=-\frac{\sqrt{65}}{-5}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{65}}{2*-5}=\frac{0+2\sqrt{65}}{-10}
=\frac{2\sqrt{65}}{-10}
=\frac{\sqrt{65}}{-5}
$