x+2x^2+x+20x=0

Solution for x+2x^2+x+20x=0 equation:

x+2x^2+x+20x=0

We add all the numbers together, and all the variables

2x^2+22x=0

a = 2; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·2·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*2}=\frac{-44}{4}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*2}=\frac{0}{4}
=0
$