x+(x+1)(x+2)=126

Solution for x+(x+1)(x+2)=126 equation:

x+(x+1)(x+2)=126

We move all terms to the left:

x+(x+1)(x+2)-(126)=0

We multiply parentheses ..

(+x^2+2x+x+2)+x-126=0

We get rid of parentheses

x^2+2x+x+x+2-126=0

We add all the numbers together, and all the variables

x^2+4x-124=0

a = 1; b = 4; c = -124;
Δ = b2-4ac
Δ = 42-4·1·(-124)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16\sqrt{2}}{2*1}=\frac{-4-16\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16\sqrt{2}}{2*1}=\frac{-4+16\sqrt{2}}{2}
$