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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x-1)(x-5)

We multiply parentheses ..

(+x^2-5x-1x+5)

We get rid of parentheses

x^2-5x-1x+5

We add all the numbers together, and all the variables

x^2-6x+5

Back to the equation:

-(x^2-6x+5)


We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

(+x^2-5x-1x+5)+7x-7=0

We get rid of parentheses

x^2-5x-1x+7x+5-7=0

We add all the numbers together, and all the variables

x^2+x-2=0

a = 1; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*1}=\frac{2}{2}
=1
$