2x^2+128x+0=x

Solution for 2x^2+128x+0=x equation:

2x^2+128x+0=x

We move all terms to the left:

2x^2+128x+0-(x)=0

We add all the numbers together, and all the variables

2x^2+127x=0

a = 2; b = 127; c = 0;
Δ = b2-4ac
Δ = 1272-4·2·0
Δ = 16129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16129}=127$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(127)-127}{2*2}=\frac{-254}{4}
=-63+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(127)+127}{2*2}=\frac{0}{4}
=0
$