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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(3x+1)(x-2)

We multiply parentheses ..

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

We get rid of parentheses

3x^2-6x+x-2

We add all the numbers together, and all the variables

3x^2-5x-2

Back to the equation:

-(3x^2-5x-2)


We get rid of parentheses

2x^2-3x^2-2x+x+5x-1+2=0

We add all the numbers together, and all the variables

-1x^2+4x+1=0

a = -1; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-1)·1
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{5}}{2*-1}=\frac{-4-2\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{5}}{2*-1}=\frac{-4+2\sqrt{5}}{-2}
$