b2+8b-39=-6

Solution for b2+8b-39=-6 equation:

b2+8b-39=-6

We move all terms to the left:

b2+8b-39-(-6)=0

We add all the numbers together, and all the variables

b^2+8b-33=0

a = 1; b = 8; c = -33;
Δ = b2-4ac
Δ = 82-4·1·(-33)
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{-(8)-14}{2*1}=\frac{-22}{2}
=-11
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*1}=\frac{6}{2}
=3
$