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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(-2x(x-3)+3x), so:

-2x(x-3)+3x

We add all the numbers together, and all the variables

3x-2x(x-3)

We multiply parentheses

-2x^2+3x+6x

We add all the numbers together, and all the variables

-2x^2+9x

Back to the equation:

-(-2x^2+9x)


We add all the numbers together, and all the variables

-(-2x^2+9x)+x+6=0

We get rid of parentheses

2x^2-9x+x+6=0

We add all the numbers together, and all the variables

2x^2-8x+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$