(x+2)(x-3)=3(x+1)-9

Solution for (x+2)(x-3)=3(x+1)-9 equation:

(x+2)(x-3)=3(x+1)-9

We move all terms to the left:

(x+2)(x-3)-(3(x+1)-9)=0

We multiply parentheses ..

(+x^2-3x+2x-6)-(3(x+1)-9)=0


We calculate terms in parentheses: -(3(x+1)-9), so:

3(x+1)-9

We multiply parentheses

3x+3-9

We add all the numbers together, and all the variables

3x-6

Back to the equation:

-(3x-6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-4x=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
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{-(-4)-4}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$