2(4k^2)+8=40

Solution for 2(4k^2)+8=40 equation:

2(4k^2)+8=40

We move all terms to the left:

2(4k^2)+8-(40)=0

We add all the numbers together, and all the variables

24k^2-32=0

a = 24; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·24·(-32)
Δ = 3072
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:

$\sqrt{\Delta}=\sqrt{3072}=\sqrt{1024*3}=\sqrt{1024}*\sqrt{3}=32\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{3}}{2*24}=\frac{0-32\sqrt{3}}{48}
=-\frac{32\sqrt{3}}{48}
=-\frac{2\sqrt{3}}{3}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{3}}{2*24}=\frac{0+32\sqrt{3}}{48}
=\frac{32\sqrt{3}}{48}
=\frac{2\sqrt{3}}{3}
$