2k^2=27k-8

Solution for 2k^2=27k-8 equation:

2k^2=27k-8

We move all terms to the left:

2k^2-(27k-8)=0

We get rid of parentheses

2k^2-27k+8=0

a = 2; b = -27; c = +8;
Δ = b2-4ac
Δ = -272-4·2·8
Δ = 665
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{665}}{2*2}=\frac{27-\sqrt{665}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{665}}{2*2}=\frac{27+\sqrt{665}}{4}
$