(8x-7)(3x-1)=(4x-1)(5x-3)

Solution for (8x-7)(3x-1)=(4x-1)(5x-3) equation:

(8x-7)(3x-1)=(4x-1)(5x-3)

We move all terms to the left:

(8x-7)(3x-1)-((4x-1)(5x-3))=0

We multiply parentheses ..

(+24x^2-8x-21x+7)-((4x-1)(5x-3))=0


We calculate terms in parentheses: -((4x-1)(5x-3)), so:

(4x-1)(5x-3)

We multiply parentheses ..

(+20x^2-12x-5x+3)

We get rid of parentheses

20x^2-12x-5x+3

We add all the numbers together, and all the variables

20x^2-17x+3

Back to the equation:

-(20x^2-17x+3)


We get rid of parentheses

24x^2-20x^2-8x-21x+17x+7-3=0

We add all the numbers together, and all the variables

4x^2-12x+4=0

a = 4; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·4·4
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{5}}{2*4}=\frac{12-4\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{5}}{2*4}=\frac{12+4\sqrt{5}}{8}
$