(13-n)n=-48

Solution for (13-n)n=-48 equation:

(13-n)n=-48

We move all terms to the left:

(13-n)n-(-48)=0

We add all the numbers together, and all the variables

(-1n+13)n-(-48)=0

We add all the numbers together, and all the variables

(-1n+13)n+48=0

We multiply parentheses

-1n^2+13n+48=0

a = -1; b = 13; c = +48;
Δ = b2-4ac
Δ = 132-4·(-1)·48
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*-1}=\frac{-32}{-2}
=+16
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*-1}=\frac{6}{-2}
=-3
$