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Five times of a positive integer is less than twice its square by 3. Find the integer.​

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Let x be the number then

[tex]5x = 2 {x}^{2} - 3 \\ 2 { x}^{2} - 5x - 3 = 0 \\ 2 {x}^{2} - 6x + x - 3 = 0 \\ 2 x(x - 3) + 1( x - 3) = 0[/tex]

or

[tex](2 \times + 1)( \times - 3) = 0[/tex]

x=3 as x is a positive integer

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