(3x-1)(x-3)-(4x-5)(3x-4)=6(x-7)-(3x-5)

Solution for (3x-1)(x-3)-(4x-5)(3x-4)=6(x-7)-(3x-5) equation:

(3x-1)(x-3)-(4x-5)(3x-4)=6(x-7)-(3x-5)

We move all terms to the left:

(3x-1)(x-3)-(4x-5)(3x-4)-(6(x-7)-(3x-5))=0

We multiply parentheses ..

(+3x^2-9x-1x+3)-(4x-5)(3x-4)-(6(x-7)-(3x-5))=0


We calculate terms in parentheses: -(6(x-7)-(3x-5)), so:

6(x-7)-(3x-5)

We multiply parentheses

6x-(3x-5)-42

We get rid of parentheses

6x-3x+5-42

We add all the numbers together, and all the variables

3x-37

Back to the equation:

-(3x-37)


We get rid of parentheses

3x^2-9x-1x-(4x-5)(3x-4)-3x+3+37=0

We multiply parentheses ..

3x^2-(+12x^2-16x-15x+20)-9x-1x-3x+3+37=0

We add all the numbers together, and all the variables

3x^2-(+12x^2-16x-15x+20)-13x+40=0

We get rid of parentheses

3x^2-12x^2+16x+15x-13x-20+40=0

We add all the numbers together, and all the variables

-9x^2+18x+20=0

a = -9; b = 18; c = +20;
Δ = b2-4ac
Δ = 182-4·(-9)·20
Δ = 1044
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:

$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{29}}{2*-9}=\frac{-18-6\sqrt{29}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{29}}{2*-9}=\frac{-18+6\sqrt{29}}{-18}
$