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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-7x+10=0

a = 1; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·1·10
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{-(-7)-3}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{10}{2}
=5
$