# Proving equation has exactly two real solutions

How can I prove that an equation has exactly 2 real solutions using some kind of real-analysis methods?

For example I have to prove it for $$x^{16}+7x^2-5=0$$.

There is a zero of derivative between any two solutions. As $$(x^{16} + 7x^2 - 5)^\prime = 16x^{15} + 14x = x\cdot(16x^{14} + 14)$$, the only zero of derivative on real line is at $$x = 0$$, so the equation has at most two solutions (would there be at least three derivative would have at least two zeros).
Hint: Take the derivative $$f'$$