x(x+2)+(x+4)+(x+6)=4x+12=212

Solution for x(x+2)+(x+4)+(x+6)=4x+12=212 equation:

x(x+2)+(x+4)+(x+6)=4x+12=212

We move all terms to the left:

x(x+2)+(x+4)+(x+6)-(4x+12)=0

We multiply parentheses

x^2+2x+(x+4)+(x+6)-(4x+12)=0

We get rid of parentheses

x^2+2x+x+x-4x+4+6-12=0

We add all the numbers together, and all the variables

x^2-2=0

a = 1; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·1·(-2)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*1}=\frac{0-2\sqrt{2}}{2}
=-\frac{2\sqrt{2}}{2}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*1}=\frac{0+2\sqrt{2}}{2}
=\frac{2\sqrt{2}}{2}
=\sqrt{2}
$