3x^2+20x=12

Solution for 3x^2+20x=12 equation:

3x^2+20x=12

We move all terms to the left:

3x^2+20x-(12)=0

a = 3; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·3·(-12)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{34}}{2*3}=\frac{-20-4\sqrt{34}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{34}}{2*3}=\frac{-20+4\sqrt{34}}{6}
$