b^2+12b=-27

Solution for b^2+12b=-27 equation:

b^2+12b=-27

We move all terms to the left:

b^2+12b-(-27)=0

We add all the numbers together, and all the variables

b^2+12b+27=0

a = 1; b = 12; c = +27;
Δ = b2-4ac
Δ = 122-4·1·27
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*1}=\frac{-18}{2}
=-9
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*1}=\frac{-6}{2}
=-3
$