n2-14n+42=-6

Solution for n2-14n+42=-6 equation:

n2-14n+42=-6

We move all terms to the left:

n2-14n+42-(-6)=0

We add all the numbers together, and all the variables

n^2-14n+48=0

a = 1; b = -14; c = +48;
Δ = b2-4ac
Δ = -142-4·1·48
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{-(-14)-2}{2*1}=\frac{12}{2}
=6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*1}=\frac{16}{2}
=8
$