n(n=14)=-33

Solution for n(n=14)=-33 equation:

n(n=14)=-33

We move all terms to the left:

n(n-(14))=0

We multiply parentheses

n^2-14n=0

a = 1; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·1·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*1}=\frac{0}{2}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*1}=\frac{28}{2}
=14
$