(x-3)(x-3)-x(x-2)=x^2+4

Solution for (x-3)(x-3)-x(x-2)=x^2+4 equation:

(x-3)(x-3)-x(x-2)=x^2+4

We move all terms to the left:

(x-3)(x-3)-x(x-2)-(x^2+4)=0

We multiply parentheses

-x^2+(x-3)(x-3)+2x-(x^2+4)=0

We get rid of parentheses

-x^2-x^2+(x-3)(x-3)+2x-4=0

We multiply parentheses ..

-x^2-x^2+(+x^2-3x-3x+9)+2x-4=0

We add all the numbers together, and all the variables

-2x^2+(+x^2-3x-3x+9)+2x-4=0

We get rid of parentheses

-2x^2+x^2-3x-3x+2x+9-4=0

We add all the numbers together, and all the variables

-1x^2-4x+5=0

a = -1; b = -4; c = +5;
Δ = b2-4ac
Δ = -42-4·(-1)·5
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*-1}=\frac{10}{-2}
=-5
$