2x+(4x+8)2x+(4x+8)=88

Solution for 2x+(4x+8)2x+(4x+8)=88 equation:

2x+(4x+8)2x+(4x+8)=88

We move all terms to the left:

2x+(4x+8)2x+(4x+8)-(88)=0

We multiply parentheses

8x^2+2x+16x+(4x+8)-88=0

We get rid of parentheses

8x^2+2x+16x+4x+8-88=0

We add all the numbers together, and all the variables

8x^2+22x-80=0

a = 8; b = 22; c = -80;
Δ = b2-4ac
Δ = 222-4·8·(-80)
Δ = 3044
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:

$\sqrt{\Delta}=\sqrt{3044}=\sqrt{4*761}=\sqrt{4}*\sqrt{761}=2\sqrt{761}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{761}}{2*8}=\frac{-22-2\sqrt{761}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{761}}{2*8}=\frac{-22+2\sqrt{761}}{16}
$