b^2+6b-48=-8

Solution for b^2+6b-48=-8 equation:

b^2+6b-48=-8

We move all terms to the left:

b^2+6b-48-(-8)=0

We add all the numbers together, and all the variables

b^2+6b-40=0

a = 1; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·1·(-40)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*1}=\frac{-20}{2}
=-10
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*1}=\frac{8}{2}
=4
$