10x-10+2x-2=(5x-5)(5x-5)

Solution for 10x-10+2x-2=(5x-5)(5x-5) equation:

10x-10+2x-2=(5x-5)(5x-5)

We move all terms to the left:

10x-10+2x-2-((5x-5)(5x-5))=0

We add all the numbers together, and all the variables

12x-((5x-5)(5x-5))-12=0

We multiply parentheses ..

-((+25x^2-25x-25x+25))+12x-12=0


We calculate terms in parentheses: -((+25x^2-25x-25x+25)), so:

(+25x^2-25x-25x+25)

We get rid of parentheses

25x^2-25x-25x+25

We add all the numbers together, and all the variables

25x^2-50x+25

Back to the equation:

-(25x^2-50x+25)


We add all the numbers together, and all the variables

12x-(25x^2-50x+25)-12=0

We get rid of parentheses

-25x^2+12x+50x-25-12=0

We add all the numbers together, and all the variables

-25x^2+62x-37=0

a = -25; b = 62; c = -37;
Δ = b2-4ac
Δ = 622-4·(-25)·(-37)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-12}{2*-25}=\frac{-74}{-50}
=1+12/25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+12}{2*-25}=\frac{-50}{-50}
=1
$