2304+b^2=3600

Solution for 2304+b^2=3600 equation:

2304+b^2=3600

We move all terms to the left:

2304+b^2-(3600)=0

We add all the numbers together, and all the variables

b^2-1296=0

a = 1; b = 0; c = -1296;
Δ = b2-4ac
Δ = 02-4·1·(-1296)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72}{2*1}=\frac{-72}{2}
=-36
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72}{2*1}=\frac{72}{2}
=36
$