3(2x+2)-3x=6+3x2

Solution for 3(2x+2)-3x=6+3x2 equation:

3(2x+2)-3x=6+3x^2

We move all terms to the left:

3(2x+2)-3x-(6+3x^2)=0

We add all the numbers together, and all the variables

-(6+3x^2)-3x+3(2x+2)=0

We multiply parentheses

-(6+3x^2)-3x+6x+6=0

We get rid of parentheses

-3x^2-3x+6x-6+6=0

We add all the numbers together, and all the variables

-3x^2+3x=0

a = -3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-3}=\frac{0}{-6}
=0
$