3x^2+12x=32

Solution for 3x^2+12x=32 equation:

3x^2+12x=32

We move all terms to the left:

3x^2+12x-(32)=0

a = 3; b = 12; c = -32;
Δ = b2-4ac
Δ = 122-4·3·(-32)
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*3}=\frac{-12-4\sqrt{33}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{33}}{2*3}=\frac{-12+4\sqrt{33}}{6}
$