2n^2=n+18

Solution for 2n^2=n+18 equation:

2n^2=n+18

We move all terms to the left:

2n^2-(n+18)=0

We get rid of parentheses

2n^2-n-18=0

We add all the numbers together, and all the variables

2n^2-1n-18=0

a = 2; b = -1; c = -18;
Δ = b2-4ac
Δ = -12-4·2·(-18)
Δ = 145
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{145}}{2*2}=\frac{1-\sqrt{145}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{145}}{2*2}=\frac{1+\sqrt{145}}{4}
$