2x^2+18x^2+40x=0

Solution for 2x^2+18x^2+40x=0 equation:

2x^2+18x^2+40x=0

We add all the numbers together, and all the variables

20x^2+40x=0

a = 20; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·20·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*20}=\frac{-80}{40}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*20}=\frac{0}{40}
=0
$