(n*n)+n-48=0

Solution for (n*n)+n-48=0 equation:

(n*n)+n-48=0

We add all the numbers together, and all the variables

(+n*n)+n-48=0

We add all the numbers together, and all the variables

n+(+n*n)-48=0

We get rid of parentheses

n+n*n-48=0

Wy multiply elements

n^2+n-48=0

a = 1; b = 1; c = -48;
Δ = b2-4ac
Δ = 12-4·1·(-48)
Δ = 193
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{-(1)-\sqrt{193}}{2*1}=\frac{-1-\sqrt{193}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*1}=\frac{-1+\sqrt{193}}{2}
$