4k^2+k=18

Solution for 4k^2+k=18 equation:

4k^2+k=18

We move all terms to the left:

4k^2+k-(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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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