n^2+20n+98=0

Solution for n^2+20n+98=0 equation:

n^2+20n+98=0

a = 1; b = 20; c = +98;
Δ = b2-4ac
Δ = 202-4·1·98
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{2}}{2*1}=\frac{-20-2\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{2}}{2*1}=\frac{-20+2\sqrt{2}}{2}
$