\(5x^3+4x-2=0\\\\
(5x^3+4x-2)'=15x^2+4\\\\
15x^2+4=0\\
15x^2=-4\\
x^2=-\dfrac{4}{15}\\
x\in\emptyset\)
The derivative of \(5x^3+4x-2\) is positive for all real numbers, which means the function is increasing in its all domain and therefore there is only one intersection with the x-axis ⇒ one solution.
