b^2+12b-22464=0

Solution for b^2+12b-22464=0 equation:

b^2+12b-22464=0

a = 1; b = 12; c = -22464;
Δ = b2-4ac
Δ = 122-4·1·(-22464)
Δ = 90000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{90000}=300$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-300}{2*1}=\frac{-312}{2}
=-156
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+300}{2*1}=\frac{288}{2}
=144
$