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

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

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

We move all terms to the left:

3x^2+20x-12-(-2x+4)=0

We get rid of parentheses

3x^2+20x+2x-4-12=0

We add all the numbers together, and all the variables

3x^2+22x-16=0

a = 3; b = 22; c = -16;
Δ = b2-4ac
Δ = 222-4·3·(-16)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*3}=\frac{-48}{6}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*3}=\frac{4}{6}
=2/3
$