b^2+18b+72=-8

Solution for b^2+18b+72=-8 equation:

b^2+18b+72=-8

We move all terms to the left:

b^2+18b+72-(-8)=0

We add all the numbers together, and all the variables

b^2+18b+80=0

a = 1; b = 18; c = +80;
Δ = b2-4ac
Δ = 182-4·1·80
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*1}=\frac{-20}{2}
=-10
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*1}=\frac{-16}{2}
=-8
$