2k^2-18=5k

Solution for 2k^2-18=5k equation:

2k^2-18=5k

We move all terms to the left:

2k^2-18-(5k)=0

a = 2; b = -5; c = -18;
Δ = b2-4ac
Δ = -52-4·2·(-18)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*2}=\frac{-8}{4}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*2}=\frac{18}{4}
=4+1/2
$