2n^2-18n-224=0

Solution for 2n^2-18n-224=0 equation:

2n^2-18n-224=0

a = 2; b = -18; c = -224;
Δ = b2-4ac
Δ = -182-4·2·(-224)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-46}{2*2}=\frac{-28}{4}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+46}{2*2}=\frac{64}{4}
=16
$