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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2x^2-4x+x-2)-((2x+1)(2x-3))=0


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

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

We multiply parentheses ..

(+4x^2-6x+2x-3)

We get rid of parentheses

4x^2-6x+2x-3

We add all the numbers together, and all the variables

4x^2-4x-3

Back to the equation:

-(4x^2-4x-3)


We get rid of parentheses

2x^2-4x^2-4x+x+4x-2+3=0

We add all the numbers together, and all the variables

-2x^2+x+1=0

a = -2; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-2)·1
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{-(1)-3}{2*-2}=\frac{-4}{-4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*-2}=\frac{2}{-4}
=-1/2
$