n2+18n+21=4

Solution for n2+18n+21=4 equation:

n2+18n+21=4

We move all terms to the left:

n2+18n+21-(4)=0

We add all the numbers together, and all the variables

n^2+18n+17=0

a = 1; b = 18; c = +17;
Δ = b2-4ac
Δ = 182-4·1·17
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-16}{2*1}=\frac{-34}{2}
=-17
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+16}{2*1}=\frac{-2}{2}
=-1
$