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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+12x^2-9x-16x+12)-((2x-4)(2x-5))=0


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

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

We multiply parentheses ..

(+4x^2-10x-8x+20)

We get rid of parentheses

4x^2-10x-8x+20

We add all the numbers together, and all the variables

4x^2-18x+20

Back to the equation:

-(4x^2-18x+20)


We get rid of parentheses

12x^2-4x^2-9x-16x+18x+12-20=0

We add all the numbers together, and all the variables

8x^2-7x-8=0

a = 8; b = -7; c = -8;
Δ = b2-4ac
Δ = -72-4·8·(-8)
Δ = 305
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{305}}{2*8}=\frac{7-\sqrt{305}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{305}}{2*8}=\frac{7+\sqrt{305}}{16}
$