n^2+13n=40

Solution for n^2+13n=40 equation:

n^2+13n=40

We move all terms to the left:

n^2+13n-(40)=0

a = 1; b = 13; c = -40;
Δ = b2-4ac
Δ = 132-4·1·(-40)
Δ = 329
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{-(13)-\sqrt{329}}{2*1}=\frac{-13-\sqrt{329}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{329}}{2*1}=\frac{-13+\sqrt{329}}{2}
$