(2x^2-2x+3)-(x^2+5x-5)=1

Solution for (2x^2-2x+3)-(x^2+5x-5)=1 equation:

(2x^2-2x+3)-(x^2+5x-5)=1

We move all terms to the left:

(2x^2-2x+3)-(x^2+5x-5)-(1)=0

We get rid of parentheses

2x^2-x^2-2x-5x+3+5-1=0

We add all the numbers together, and all the variables

x^2-7x+7=0

a = 1; b = -7; c = +7;
Δ = b2-4ac
Δ = -72-4·1·7
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{21}}{2*1}=\frac{7-\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{21}}{2*1}=\frac{7+\sqrt{21}}{2}
$