X^2+4x+4=6x-12+16

Solution for X^2+4x+4=6x-12+16 equation:

X^2+4X+4=6X-12+16

We move all terms to the left:

X^2+4X+4-(6X-12+16)=0

We add all the numbers together, and all the variables

X^2+4X-(6X+4)+4=0

We get rid of parentheses

X^2+4X-6X-4+4=0

We add all the numbers together, and all the variables

X^2-2X=0

a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2}
=0
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2}
=2
$