n2+n-462=0

Solution for n2+n-462=0 equation:

n2+n-462=0

We add all the numbers together, and all the variables

n^2+n-462=0

a = 1; b = 1; c = -462;
Δ = b2-4ac
Δ = 12-4·1·(-462)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-43}{2*1}=\frac{-44}{2}
=-22
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+43}{2*1}=\frac{42}{2}
=21
$