2x.4(x+1)=128

Solution for 2x.4(x+1)=128 equation:

2x.4(x+1)=128

We move all terms to the left:

2x.4(x+1)-(128)=0

We multiply parentheses

2x^2+2x-128=0

a = 2; b = 2; c = -128;
Δ = b2-4ac
Δ = 22-4·2·(-128)
Δ = 1028
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:

$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{257}}{2*2}=\frac{-2-2\sqrt{257}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{257}}{2*2}=\frac{-2+2\sqrt{257}}{4}
$