n^2+n-4n=0

Solution for n^2+n-4n=0 equation:

n^2+n-4n=0

We add all the numbers together, and all the variables

n^2-3n=0

a = 1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·1·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*1}=\frac{0}{2}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*1}=\frac{6}{2}
=3
$