2k^2+60=22k

Solution for 2k^2+60=22k equation:

2k^2+60=22k

We move all terms to the left:

2k^2+60-(22k)=0

a = 2; b = -22; c = +60;
Δ = b2-4ac
Δ = -222-4·2·60
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2}{2*2}=\frac{20}{4}
=5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2}{2*2}=\frac{24}{4}
=6
$