Given:

$\begin{array}{l}
16 x^{2}+24 x+1 \\
\Rightarrow 16 x^{2}-24 x-1=0
\end{array}$

On comparing it with $a x^{2}+b x+x=0$, we get; $a=16, b=-24$ and $c=-1$

Discriminant $D$ is given by:

$\begin{array}{l}
D=\left(b^{2}-4 a c\right) \\
=(-24)^{2}-4 \times 16 \times(-1)
\end{array}$

$\begin{array}{l}
=576+(64) \\
=640>0
\end{array}$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by:

$\begin{array}{l}
\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-24)+\sqrt{640}}{2 \times 16}=\frac{24+8 \sqrt{10}}{32}=\frac{8(3+\sqrt{10})}{32}=\frac{(3+\sqrt{10})}{4} \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-24)-\sqrt{640}}{2 \times 16}=\frac{24-8 \sqrt{10}}{32}=\frac{8(3-\sqrt{10})}{32}=\frac{(3-\sqrt{10})}{4}
\end{array}$

Thus, the roots of the equation are $\frac{(3+\sqrt{10})}{4}$ and $\frac{(3-\sqrt{10})}{4}$.