P(n)=n^2+10n+5

Solution for P(n)=n^2+10n+5 equation:

(P)=P^2+10P+5

We move all terms to the left:

(P)-(P^2+10P+5)=0

We get rid of parentheses

-P^2+P-10P-5=0

We add all the numbers together, and all the variables

-1P^2-9P-5=0

a = -1; b = -9; c = -5;
Δ = b2-4ac
Δ = -92-4·(-1)·(-5)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{61}}{2*-1}=\frac{9-\sqrt{61}}{-2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{61}}{2*-1}=\frac{9+\sqrt{61}}{-2}
$