Solve (x-4)² + x(x-12) = 16: Finding Parameter x

Let's open the parentheses, remembering that there might be more than one solution for the value of X:

$(x-4)^2+x(x-12)=16$

$x^2-8x+16+x^2-12x=16$

$2x^2-20x=0$

$2x(x-10)=0$

Therefore:

$x-10=0$

$x=10$

Or:

$2x=0$

$x=0$