n2+42=-13n

Solution for n2+42=-13n equation:

n2+42=-13n

We move all terms to the left:

n2+42-(-13n)=0

We add all the numbers together, and all the variables

n^2-(-13n)+42=0

We get rid of parentheses

n^2+13n+42=0

a = 1; b = 13; c = +42;
Δ = b2-4ac
Δ = 132-4·1·42
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*1}=\frac{-14}{2}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*1}=\frac{-12}{2}
=-6
$