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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-1)·2
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*-1}=\frac{-2-2\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*-1}=\frac{-2+2\sqrt{3}}{-2}
$