n^2+11n+25=n+5

Solution for n^2+11n+25=n+5 equation:

n^2+11n+25=n+5

We move all terms to the left:

n^2+11n+25-(n+5)=0

We get rid of parentheses

n^2+11n-n-5+25=0

We add all the numbers together, and all the variables

n^2+10n+20=0

a = 1; b = 10; c = +20;
Δ = b2-4ac
Δ = 102-4·1·20
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*1}=\frac{-10-2\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*1}=\frac{-10+2\sqrt{5}}{2}
$