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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

3(2x-1)

We multiply parentheses

6x-3

Back to the equation:

-(6x-3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-7x=0

a = 2; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2}=\frac{14}{4}
=3+1/2
$