8k(6k-4)=10

Solution for 8k(6k-4)=10 equation:

8k(6k-4)=10

We move all terms to the left:

8k(6k-4)-(10)=0

We multiply parentheses

48k^2-32k-10=0

a = 48; b = -32; c = -10;
Δ = b2-4ac
Δ = -322-4·48·(-10)
Δ = 2944
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:

$\sqrt{\Delta}=\sqrt{2944}=\sqrt{64*46}=\sqrt{64}*\sqrt{46}=8\sqrt{46}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{46}}{2*48}=\frac{32-8\sqrt{46}}{96}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{46}}{2*48}=\frac{32+8\sqrt{46}}{96}
$