4n^2-n-18=0

Solution for 4n^2-n-18=0 equation:

4n^2-n-18=0

We add all the numbers together, and all the variables

4n^2-1n-18=0

a = 4; b = -1; c = -18;
Δ = b2-4ac
Δ = -12-4·4·(-18)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*4}=\frac{-16}{8}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*4}=\frac{18}{8}
=2+1/4
$