n^2=325

Solution for n^2=325 equation:

n^2=325

We move all terms to the left:

n^2-(325)=0

a = 1; b = 0; c = -325;
Δ = b2-4ac
Δ = 02-4·1·(-325)
Δ = 1300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{13}}{2*1}=\frac{0-10\sqrt{13}}{2}
=-\frac{10\sqrt{13}}{2}
=-5\sqrt{13}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{13}}{2*1}=\frac{0+10\sqrt{13}}{2}
=\frac{10\sqrt{13}}{2}
=5\sqrt{13}
$