b^2+6b=36

Solution for b^2+6b=36 equation:

b^2+6b=36

We move all terms to the left:

b^2+6b-(36)=0

a = 1; b = 6; c = -36;
Δ = b2-4ac
Δ = 62-4·1·(-36)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{5}}{2*1}=\frac{-6-6\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{5}}{2*1}=\frac{-6+6\sqrt{5}}{2}
$