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

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

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

We multiply parentheses ..

(+6x^2-5x-6x+5)+(x-1)(x+2)=0

We get rid of parentheses

6x^2-5x-6x+(x-1)(x+2)+5=0

We multiply parentheses ..

6x^2+(+x^2+2x-1x-2)-5x-6x+5=0

We add all the numbers together, and all the variables

6x^2+(+x^2+2x-1x-2)-11x+5=0

We get rid of parentheses

6x^2+x^2+2x-1x-11x-2+5=0

We add all the numbers together, and all the variables

7x^2-10x+3=0

a = 7; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·7·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2*7}=\frac{6}{14}
=3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2*7}=\frac{14}{14}
=1
$