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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2-2x+3x-6)), so:

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

We get rid of parentheses

x^2-2x+3x-6

We add all the numbers together, and all the variables

x^2+x-6

Back to the equation:

-(x^2+x-6)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-3x=0

a = -1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-1)·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*-1}=\frac{0}{-2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-1}=\frac{6}{-2}
=-3
$