(X-3)(x-3)+(x+1)(x+1)=12

Solution for (X-3)(x-3)+(x+1)(x+1)=12 equation:

(X-3)(X-3)+(X+1)(X+1)=12

We move all terms to the left:

(X-3)(X-3)+(X+1)(X+1)-(12)=0

We multiply parentheses ..

(+X^2-3X-3X+9)+(X+1)(X+1)-12=0

We get rid of parentheses

X^2-3X-3X+(X+1)(X+1)+9-12=0

We multiply parentheses ..

X^2+(+X^2+X+X+1)-3X-3X+9-12=0

We add all the numbers together, and all the variables

X^2+(+X^2+X+X+1)-6X-3=0

We get rid of parentheses

X^2+X^2+X+X-6X+1-3=0

We add all the numbers together, and all the variables

2X^2-4X-2=0

a = 2; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·2·(-2)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{2}}{2*2}=\frac{4-4\sqrt{2}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2}}{2*2}=\frac{4+4\sqrt{2}}{4}
$