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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x+1)(x+1)-3

We multiply parentheses ..

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

We get rid of parentheses

x^2+x+x+1-3

We add all the numbers together, and all the variables

x^2+2x-2

Back to the equation:

-(x^2+2x-2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-3x=0

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