(x+2)(x-3)+2x=x2-3(x-1)(x+2)

Solution for (x+2)(x-3)+2x=x2-3(x-1)(x+2) equation:

(x+2)(x-3)+2x=x2-3(x-1)(x+2)

We move all terms to the left:

(x+2)(x-3)+2x-(x2-3(x-1)(x+2))=0

We add all the numbers together, and all the variables

2x+(x+2)(x-3)-(x2-3(x-1)(x+2))=0

We multiply parentheses ..

(+x^2-3x+2x-6)+2x-(x2-3(x-1)(x+2))=0


We calculate terms in parentheses: -(x2-3(x-1)(x+2)), so:

x2-3(x-1)(x+2)

We add all the numbers together, and all the variables

x^2-3(x-1)(x+2)

We multiply parentheses ..

x^2-3(+x^2+2x-1x-2)

We multiply parentheses

x^2-3x^2-6x+3x+6

We add all the numbers together, and all the variables

-2x^2-3x+6

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2+4x-12=0

a = 3; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·3·(-12)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*3}=\frac{-4-4\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*3}=\frac{-4+4\sqrt{10}}{6}
$