b^2-23b+138=12

Solution for b^2-23b+138=12 equation:

b^2-23b+138=12

We move all terms to the left:

b^2-23b+138-(12)=0

We add all the numbers together, and all the variables

b^2-23b+126=0

a = 1; b = -23; c = +126;
Δ = b2-4ac
Δ = -232-4·1·126
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-5}{2*1}=\frac{18}{2}
=9
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+5}{2*1}=\frac{28}{2}
=14
$