4x^2+12x=24

Solution for 4x^2+12x=24 equation:

4x^2+12x=24

We move all terms to the left:

4x^2+12x-(24)=0

a = 4; b = 12; c = -24;
Δ = b2-4ac
Δ = 122-4·4·(-24)
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{33}}{2*4}=\frac{-12-4\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{33}}{2*4}=\frac{-12+4\sqrt{33}}{8}
$