Five times of a positive integer is 3 less than twice the square of that number. The number is -

Solution (By Examveda Team)

$$\eqalign{ & {\text{Let }}x{\text{ is the number}} \cr & {\text{According to question,}} \cr & 5x = 2{x^2} - 3 \cr & 2{x^2} - 5x - 3 = 0 \cr & 2{x^2} - 6x + x - 3 = 0 \cr & 2x\left( {x - 3} \right) + 1\left( {x - 3} \right) = 0 \cr & \left( {2x + 1} \right)\left( {x - 3} \right) = 0 \cr & x = 3 \cr} $$