x+x(x+2)+(x+4)=54

Solution for x+x(x+2)+(x+4)=54 equation:

x+x(x+2)+(x+4)=54

We move all terms to the left:

x+x(x+2)+(x+4)-(54)=0

We multiply parentheses

x^2+x+2x+(x+4)-54=0

We get rid of parentheses

x^2+x+2x+x+4-54=0

We add all the numbers together, and all the variables

x^2+4x-50=0

a = 1; b = 4; c = -50;
Δ = b2-4ac
Δ = 42-4·1·(-50)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{6}}{2*1}=\frac{-4-6\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{6}}{2*1}=\frac{-4+6\sqrt{6}}{2}
$