b^2-14b+8=-10

Solution for b^2-14b+8=-10 equation:

b^2-14b+8=-10

We move all terms to the left:

b^2-14b+8-(-10)=0

We add all the numbers together, and all the variables

b^2-14b+18=0

a = 1; b = -14; c = +18;
Δ = b2-4ac
Δ = -142-4·1·18
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{31}}{2*1}=\frac{14-2\sqrt{31}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{31}}{2*1}=\frac{14+2\sqrt{31}}{2}
$