a^2+12a=24

Solution for a^2+12a=24 equation:

a^2+12a=24

We move all terms to the left:

a^2+12a-(24)=0

a = 1; b = 12; c = -24;
Δ = b2-4ac
Δ = 122-4·1·(-24)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{15}}{2*1}=\frac{-12-4\sqrt{15}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{15}}{2*1}=\frac{-12+4\sqrt{15}}{2}
$