16k^2-24=0

Solution for 16k^2-24=0 equation:

16k^2-24=0

a = 16; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·16·(-24)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*16}=\frac{0-16\sqrt{6}}{32}
=-\frac{16\sqrt{6}}{32}
=-\frac{\sqrt{6}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*16}=\frac{0+16\sqrt{6}}{32}
=\frac{16\sqrt{6}}{32}
=\frac{\sqrt{6}}{2}
$