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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

2(x-3)+x

We add all the numbers together, and all the variables

x+2(x-3)

We multiply parentheses

x+2x-6

We add all the numbers together, and all the variables

3x-6

Back to the equation:

-(3x-6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2-9x+6=0

a = 3; b = -9; c = +6;
Δ = b2-4ac
Δ = -92-4·3·6
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{-(-9)-3}{2*3}=\frac{6}{6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*3}=\frac{12}{6}
=2
$