b^2+9b+7=-6

Solution for b^2+9b+7=-6 equation:

b^2+9b+7=-6

We move all terms to the left:

b^2+9b+7-(-6)=0

We add all the numbers together, and all the variables

b^2+9b+13=0

a = 1; b = 9; c = +13;
Δ = b2-4ac
Δ = 92-4·1·13
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{29}}{2*1}=\frac{-9-\sqrt{29}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{29}}{2*1}=\frac{-9+\sqrt{29}}{2}
$