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

Solution for (5x-1)(3x-2)+(6x-4)(x-1)=0 equation:

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

We multiply parentheses ..

(+15x^2-10x-3x+2)+(6x-4)(x-1)=0

We get rid of parentheses

15x^2-10x-3x+(6x-4)(x-1)+2=0

We multiply parentheses ..

15x^2+(+6x^2-6x-4x+4)-10x-3x+2=0

We add all the numbers together, and all the variables

15x^2+(+6x^2-6x-4x+4)-13x+2=0

We get rid of parentheses

15x^2+6x^2-6x-4x-13x+4+2=0

We add all the numbers together, and all the variables

21x^2-23x+6=0

a = 21; b = -23; c = +6;
Δ = b2-4ac
Δ = -232-4·21·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-5}{2*21}=\frac{18}{42}
=3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+5}{2*21}=\frac{28}{42}
=2/3
$