2n^-20n^2=0

Solution for 2n^-20n^2=0 equation:

2n^-20n^2=0

We add all the numbers together, and all the variables

-20n^2+2n=0

a = -20; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-20)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-20}=\frac{-4}{-40}
=1/10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-20}=\frac{0}{-40}
=0
$