8(2k+1+3k)k=2

Solution for 8(2k+1+3k)k=2 equation:

8(2k+1+3k)k=2

We move all terms to the left:

8(2k+1+3k)k-(2)=0

We add all the numbers together, and all the variables

8(5k+1)k-2=0

We multiply parentheses

40k^2+8k-2=0

a = 40; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·40·(-2)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{6}}{2*40}=\frac{-8-8\sqrt{6}}{80}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{6}}{2*40}=\frac{-8+8\sqrt{6}}{80}
$