2k+4k^2+6k^2+k=1

Solution for 2k+4k^2+6k^2+k=1 equation:

2k+4k^2+6k^2+k=1

We move all terms to the left:

2k+4k^2+6k^2+k-(1)=0

We add all the numbers together, and all the variables

10k^2+3k-1=0

a = 10; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·10·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*10}=\frac{-10}{20}
=-1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*10}=\frac{4}{20}
=1/5
$