n^2-n-4=0

Solution for n^2-n-4=0 equation:

n^2-n-4=0

We add all the numbers together, and all the variables

n^2-1n-4=0

a = 1; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·1·(-4)
Δ = 17
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{17}}{2*1}=\frac{1-\sqrt{17}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{17}}{2*1}=\frac{1+\sqrt{17}}{2}
$