b^2-23b+132=0

Solution for b^2-23b+132=0 equation:

b^2-23b+132=0

a = 1; b = -23; c = +132;
Δ = b2-4ac
Δ = -232-4·1·132
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-1}{2*1}=\frac{22}{2}
=11
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+1}{2*1}=\frac{24}{2}
=12
$