k+(k+1)+(k+2)k+(k+1)+(k+2)=4K

Solution for k+(k+1)+(k+2)k+(k+1)+(k+2)=4K equation:

k+(k+1)+(k+2)k+(k+1)+(k+2)=4

We move all terms to the left:

k+(k+1)+(k+2)k+(k+1)+(k+2)-(4)=0

We multiply parentheses

k^2+k+(k+1)+2k+(k+1)+(k+2)-4=0

We get rid of parentheses

k^2+k+k+2k+k+k+1+1+2-4=0

We add all the numbers together, and all the variables

k^2+6k=0

a = 1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*1}=\frac{-12}{2}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*1}=\frac{0}{2}
=0
$