$16 x^{2}=24 x+1$
Given :
$16 x^{2}=24 x+1$
$\Rightarrow 16 x^{2}-24 x-1=0$
On comparing it with $a x^{2}+b x+c=0$
$a=16, b=-24$ and $c=-1$
Discriminant $D$ is given by:
$D=\left(b^{2}-4 a c\right)$
$=(-24)^{2}-4 \times 16 \times(-1)$
$=576+(64)$
$=640>0$
Hence, the roots of the equation are real,
Roots $\alpha$ and $\beta$ are given by :
$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-24)+\sqrt{640}}{2 \times 16}=\frac{24+8 \sqrt{10}}{32}=\frac{8(3+\sqrt{10})}{32}=\frac{(3+\sqrt{10})}{4}$
$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-24)-\sqrt{640}}{2 \times 16}=\frac{24-8 \sqrt{10}}{32}=\frac{8(3-\sqrt{10})}{32}=\frac{(3-\sqrt{10})}{4}$
Thus, the roots of the equation are $\frac{(3+\sqrt{10})}{4}$ and $\frac{(3-\sqrt{10})}{4}$.
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