6k(k+2)+2(2k-5)=22

Solution for 6k(k+2)+2(2k-5)=22 equation:

6k(k+2)+2(2k-5)=22

We move all terms to the left:

6k(k+2)+2(2k-5)-(22)=0

We multiply parentheses

6k^2+12k+4k-10-22=0

We add all the numbers together, and all the variables

6k^2+16k-32=0

a = 6; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·6·(-32)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*6}=\frac{-48}{12}
=-4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*6}=\frac{16}{12}
=1+1/3
$