b^2=65

Solution for b^2=65 equation:

b^2=65

We move all terms to the left:

b^2-(65)=0

a = 1; b = 0; c = -65;
Δ = b2-4ac
Δ = 02-4·1·(-65)
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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