n^2+33=-14n

Solution for n^2+33=-14n equation:

n^2+33=-14n

We move all terms to the left:

n^2+33-(-14n)=0

We get rid of parentheses

n^2+14n+33=0

a = 1; b = 14; c = +33;
Δ = b2-4ac
Δ = 142-4·1·33
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-8}{2*1}=\frac{-22}{2}
=-11
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+8}{2*1}=\frac{-6}{2}
=-3
$