3n^2-n-500=0

Solution for 3n^2-n-500=0 equation:

3n^2-n-500=0

We add all the numbers together, and all the variables

3n^2-1n-500=0

a = 3; b = -1; c = -500;
Δ = b2-4ac
Δ = -12-4·3·(-500)
Δ = 6001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{6001}}{2*3}=\frac{1-\sqrt{6001}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{6001}}{2*3}=\frac{1+\sqrt{6001}}{6}
$