(3x-2)(x+3)=(2x+9)(x-1)+7

Solution for (3x-2)(x+3)=(2x+9)(x-1)+7 equation:

(3x-2)(x+3)=(2x+9)(x-1)+7

We move all terms to the left:

(3x-2)(x+3)-((2x+9)(x-1)+7)=0

We multiply parentheses ..

(+3x^2+9x-2x-6)-((2x+9)(x-1)+7)=0


We calculate terms in parentheses: -((2x+9)(x-1)+7), so:

(2x+9)(x-1)+7

We multiply parentheses ..

(+2x^2-2x+9x-9)+7

We get rid of parentheses

2x^2-2x+9x-9+7

We add all the numbers together, and all the variables

2x^2+7x-2

Back to the equation:

-(2x^2+7x-2)


We get rid of parentheses

3x^2-2x^2+9x-2x-7x-6+2=0

We add all the numbers together, and all the variables

x^2-4=0

a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
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{-(0)-4}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2}
=2
$