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

Solution for (2-3x)+(2x-3x^2)=0 equation:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2-1x+2=0

a = -3; b = -1; c = +2;
Δ = b2-4ac
Δ = -12-4·(-3)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-3}=\frac{-4}{-6}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-3}=\frac{6}{-6}
=-1
$