Solve the equation 16x2+192x-1863.2=0 numerically using Qin Jiushao's procedure.

A polynomial equation will be solved using this approach, one unknown digit at a time. Here, start by writing the coefficients of the various powers of $x$

$$\begin{gathered} \underline{x^2 x^1 x^0} \\ 16 192 -18632 \end{gathered}$$

Identifying how many digits the answer will have is the first task. What will it be, tens or hundreds.

A cursory examination reveals that if $x = 10, 10$ is excessively large because $1600 + 1920$ is larger than $1863.2.$
The answer is therefore less than $10.$ Now since $\frac{1863}{192}$ is larger than $9$, you could try that number, but you generally won't because you usually need to try a smaller first digit. So try $8$.

$$\begin{gathered} \underline{\underline{ x^2 x^1 x^0} \\ 16 192 -1863.2 \\ 8 128 2480.0} \\ 16 310 616.8 \end{gathered}$$

However, since $616.8$is positive, $8$ was a poor approximation. Try $6$.

$$\begin{gathered} \underline{\underline{ x^2 x^1 x^0} \\ 16 192 -1863.2 \\ 6 96 1728} \\ 16 288 -135.2 \\ \underline{6 96 } \\ 16 384 \end{gathered}$$

The equation $16y^2+ 384y 135.2 = 0$ must now be solved, with $x = 6 + y$. And y is less than $10$, which is known. The initial digit of $384$;  $3$ fits into the first two digits of $135.2$; $13$, almost four times, leading us to surmise that the following digit could be a $4$. But since $384$ is close to $400$ and $4$ only enters the number $13$ three times, the following digit will likely be $3$.
Attempt it now

$$\begin{gathered} \underline{ y^{2 } y^1 y^0 } \\ 16 384 -135.2 \\ \underline{3 4.8 116.64} \\ 16 388.8 -18.56 \\ \underline{3 4.8 } \\ 16 393.6 \end{gathered}$$

The equation $16z^2 + 393.6z+18.56 = 0$ must now be solved, with $y =0.3+z$. As for the initial digit of $383.6$, which is $4$,

roughly $4$ times into $18$. Try $4$.

$$\begin{gathered} \underline{ z^2 z^1 z^0 } \\ 16 393.6 18.56 \\ \underline{0.04 0.64 15.7696 } \\ 16 394.24 -2.7904 \\ \underline{0.04 0.64 } \\ 16 394.88 \end{gathered}$$

Our current response is $6.34$. We may proceed in this way to obtain successive digits, or we could approximatively determine the following few digits by dividing $394.88$ into $2.7904$. $395$ enters $2.7904$ approximately $0.0070643$ times. $6.3470643.$