n^2-14n+16=-8-4n

Solution for n^2-14n+16=-8-4n equation:

n^2-14n+16=-8-4n

We move all terms to the left:

n^2-14n+16-(-8-4n)=0

We add all the numbers together, and all the variables

n^2-14n-(-4n-8)+16=0

We get rid of parentheses

n^2-14n+4n+8+16=0

We add all the numbers together, and all the variables

n^2-10n+24=0

a = 1; b = -10; c = +24;
Δ = b2-4ac
Δ = -102-4·1·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*1}=\frac{8}{2}
=4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*1}=\frac{12}{2}
=6
$