b^2=3600

Solution for b^2=3600 equation:

b^2=3600

We move all terms to the left:

b^2-(3600)=0

a = 1; b = 0; c = -3600;
Δ = b2-4ac
Δ = 02-4·1·(-3600)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-120}{2*1}=\frac{-120}{2}
=-60
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+120}{2*1}=\frac{120}{2}
=60
$