3a^2+12a=27

Solution for 3a^2+12a=27 equation:

3a^2+12a=27

We move all terms to the left:

3a^2+12a-(27)=0

a = 3; b = 12; c = -27;
Δ = b2-4ac
Δ = 122-4·3·(-27)
Δ = 468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{13}}{2*3}=\frac{-12-6\sqrt{13}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{13}}{2*3}=\frac{-12+6\sqrt{13}}{6}
$