Let the number be $$x$$
According to ques,
=> $$x - (5 \times \frac{1}{x}) = \frac{19}{2}$$
=> $$\frac{x^2 - 5}{x} = \frac{19}{2}$$
=> $$2x^2 - 10 = 19x$$
=> $$2x^2 - 19x - 10 = 0$$
=> $$2x^2 + x - 20x - 10 = 0$$
=> $$x(2x + 1) - 10(2x + 1) = 0$$
=> $$(2x + 1) (x - 10) = 0$$
=> $$x = \frac{-1}{2} , 10$$
Since $$x$$ can't be negative, => $$x = 10$$
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