3k^2+k=27

Solution for 3k^2+k=27 equation:

3k^2+k=27

We move all terms to the left:

3k^2+k-(27)=0

a = 3; b = 1; c = -27;
Δ = b2-4ac
Δ = 12-4·3·(-27)
Δ = 325
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5\sqrt{13}}{2*3}=\frac{-1-5\sqrt{13}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5\sqrt{13}}{2*3}=\frac{-1+5\sqrt{13}}{6}
$