b^2+8b+13=0

Solution for b^2+8b+13=0 equation:

b^2+8b+13=0

a = 1; b = 8; c = +13;
Δ = b2-4ac
Δ = 82-4·1·13
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{3}}{2*1}=\frac{-8-2\sqrt{3}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{3}}{2*1}=\frac{-8+2\sqrt{3}}{2}
$