2n^2+5n+0=0

Solution for 2n^2+5n+0=0 equation:

2n^2+5n+0=0

We add all the numbers together, and all the variables

2n^2+5n=0

a = 2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·2·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*2}=\frac{-10}{4}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*2}=\frac{0}{4}
=0
$