b^2+9b+9=0

Solution for b^2+9b+9=0 equation:

b^2+9b+9=0

a = 1; b = 9; c = +9;
Δ = b2-4ac
Δ = 92-4·1·9
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{5}}{2*1}=\frac{-9-3\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{5}}{2*1}=\frac{-9+3\sqrt{5}}{2}
$