Do I need a win streak? You have played \$N\$ matches in some game where each match can only result in one of the two outcomes: win or loss. Currently, you have \$W\$ wins. You want to have a win percentage of \$P\$ or more, playing as few matches as possible. Output the minimum win streak that you need. Assume the current win streak is at \$0\$.
For example: If \$N=10, W=2, P=50\$, then  you can win \$6\$ matches in a row, bringing your win percentage to \$\frac{2+6}{10+6} = \frac{8}{16} = 50\%\$. You cannot have a win percentage of \$50\$ or more earlier than this. So the answer for this case is \$6\$.
Examples
W, N, P -&gt;
2, 10, 50%      -&gt; 6
3, 15, 50%      -&gt; 9
35, 48, 0.75    -&gt; 4
19, 21, 0.91    -&gt; 2
9, 10, 50%      -&gt; 0
0, 1, 1/100     -&gt; 1
43, 281, 24/100 -&gt; 33
0, 6, 52%       -&gt; 7

Rules

\$N\$ and \$W\$ will be integers with \$0 \le W &lt; N\$.
The percentage \$P\$ will be an integer between \$1\$ and \$100\$ inclusive. You can also choose to take a decimal value between \$0\$ and \$1\$ inclusive instead, which will contain no more than \$2\$ decimal places, or take it as a fraction.
You can take the inputs in any convenient format.
Standard loopholes are forbidden.
This is code-golf, so the shortest code in bytes wins.
Python 3, 35 bytes

lambda w,n,p:-min(0,(w-p*n)//(1-p))

Try it online!
-4 bytes thanks to pajonk
Some simple math: if \$X\$ is the number of wins we need, then we have:
\$\frac{W+X}{N+X}\geq P\$
\$W+X\geq PN+PX\$
\$X-PX\geq PN-W\$
\$X(1-P)\geq PN-W\$
Since \$P&lt;1\$, then \$1-P&gt;0\$, so we can divide from both sides without division by zero or flipping the sign.
\$X\geq\frac{PN-W}{1-P}\$