n2+n-600=0

Solution for n2+n-600=0 equation:

n2+n-600=0

We add all the numbers together, and all the variables

n^2+n-600=0

a = 1; b = 1; c = -600;
Δ = b2-4ac
Δ = 12-4·1·(-600)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-49}{2*1}=\frac{-50}{2}
=-25
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+49}{2*1}=\frac{48}{2}
=24
$