b^2=2304

Solution for b^2=2304 equation:

b^2=2304

We move all terms to the left:

b^2-(2304)=0

a = 1; b = 0; c = -2304;
Δ = b2-4ac
Δ = 02-4·1·(-2304)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-96}{2*1}=\frac{-96}{2}
=-48
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+96}{2*1}=\frac{96}{2}
=48
$