b^2-22b+112=0

Solution for b^2-22b+112=0 equation:

b^2-22b+112=0

a = 1; b = -22; c = +112;
Δ = b2-4ac
Δ = -222-4·1·112
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-6}{2*1}=\frac{16}{2}
=8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+6}{2*1}=\frac{28}{2}
=14
$