2n^2-n-4=4

Solution for 2n^2-n-4=4 equation:

2n^2-n-4=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

2n^2-1n-8=0

a = 2; b = -1; c = -8;
Δ = b2-4ac
Δ = -12-4·2·(-8)
Δ = 65
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{-(-1)-\sqrt{65}}{2*2}=\frac{1-\sqrt{65}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*2}=\frac{1+\sqrt{65}}{4}
$