(n)=2n2+4n-16

Solution for (n)=2n2+4n-16 equation:

(n)=2n^2+4n-16

We move all terms to the left:

(n)-(2n^2+4n-16)=0

We get rid of parentheses

-2n^2+n-4n+16=0

We add all the numbers together, and all the variables

-2n^2-3n+16=0

a = -2; b = -3; c = +16;
Δ = b2-4ac
Δ = -32-4·(-2)·16
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{137}}{2*-2}=\frac{3-\sqrt{137}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{137}}{2*-2}=\frac{3+\sqrt{137}}{-4}
$