16b+b^2=132

Solution for 16b+b^2=132 equation:

16b+b^2=132

We move all terms to the left:

16b+b^2-(132)=0

a = 1; b = 16; c = -132;
Δ = b2-4ac
Δ = 162-4·1·(-132)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-28}{2*1}=\frac{-44}{2}
=-22
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+28}{2*1}=\frac{12}{2}
=6
$