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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-7x=0

a = 2; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2}=\frac{14}{4}
=3+1/2
$