3x^2=16x+12=3x^2+4x-20x-12

Solution for 3x^2=16x+12=3x^2+4x-20x-12 equation:

3x^2=16x+12=3x^2+4x-20x-12

We move all terms to the left:

3x^2-(16x+12)=0

We get rid of parentheses

3x^2-16x-12=0

a = 3; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·3·(-12)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*3}=\frac{36}{6}
=6
$