(2x-3)(x+1)=(2x-3)(4x-5)

Solution for (2x-3)(x+1)=(2x-3)(4x-5) equation:

(2x-3)(x+1)=(2x-3)(4x-5)

We move all terms to the left:

(2x-3)(x+1)-((2x-3)(4x-5))=0

We multiply parentheses ..

(+2x^2+2x-3x-3)-((2x-3)(4x-5))=0


We calculate terms in parentheses: -((2x-3)(4x-5)), so:

(2x-3)(4x-5)

We multiply parentheses ..

(+8x^2-10x-12x+15)

We get rid of parentheses

8x^2-10x-12x+15

We add all the numbers together, and all the variables

8x^2-22x+15

Back to the equation:

-(8x^2-22x+15)


We get rid of parentheses

2x^2-8x^2+2x-3x+22x-3-15=0

We add all the numbers together, and all the variables

-6x^2+21x-18=0

a = -6; b = 21; c = -18;
Δ = b2-4ac
Δ = 212-4·(-6)·(-18)
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{-(21)-3}{2*-6}=\frac{-24}{-12}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3}{2*-6}=\frac{-18}{-12}
=1+1/2
$