1. Let $f(x)=x^7+14x^5+16x^3+30x-560.$ We need the number of real solutions of $f(x)=0.$

2. Use derivatives to study monotonicity.

Differentiate: $f'(x)=7x^6+70x^4+48x^2+30.$

Now observe:

• $7x^6\ge 0$
• $70x^4\ge 0$
• $48x^2\ge 0$
• $30>0$

Hence, $f'(x)=7x^6+70x^4+48x^2+30>0 \quad \text{for every real }x.$

So $f(x)$ is strictly increasing on all real numbers.

3. A strictly increasing continuous polynomial can cross the $x$-axis at most once.

Also, $\lim_{x\to -\infty} f(x)=-\infty,$ $\lim_{x\to +\infty} f(x)=+\infty.$

Since $f$ is continuous, by the Intermediate Value Theorem it must cross the $x$-axis at least once.

Because it is strictly increasing, it can cross only once.

Therefore, the equation has exactly one real solution.

4. Check with options:

• A: $7$ ❌
• B: $1$ ✅
• C: $3$ ❌
• D: $5$ ❌

So the correct option is B.