2x=3x^2+3x

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

2x=3x^2+3x

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2-1x=0

a = -3; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-3)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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