r^2+12r=27=0

Solution for r^2+12r=27=0 equation:

r^2+12r=27=0

We move all terms to the left:

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

a = 1; b = 12; c = -27;
Δ = b2-4ac
Δ = 122-4·1·(-27)
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{7}}{2*1}=\frac{-12-6\sqrt{7}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{7}}{2*1}=\frac{-12+6\sqrt{7}}{2}
$