2x+(4x+1)=8x^2+2x

Solution for 2x+(4x+1)=8x^2+2x equation:

2x+(4x+1)=8x^2+2x

We move all terms to the left:

2x+(4x+1)-(8x^2+2x)=0

We get rid of parentheses

-8x^2+2x+4x-2x+1=0

We add all the numbers together, and all the variables

-8x^2+4x+1=0

a = -8; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-8)·1
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*-8}=\frac{-4-4\sqrt{3}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*-8}=\frac{-4+4\sqrt{3}}{-16}
$