2n^2+16n=200

Solution for 2n^2+16n=200 equation:

2n^2+16n=200

We move all terms to the left:

2n^2+16n-(200)=0

a = 2; b = 16; c = -200;
Δ = b2-4ac
Δ = 162-4·2·(-200)
Δ = 1856
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:

$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{29}}{2*2}=\frac{-16-8\sqrt{29}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{29}}{2*2}=\frac{-16+8\sqrt{29}}{4}
$