Five times a positive integer is 3 less than twice the square of that integer. Form the required quadratic equation to find the integer, and then solve the equation to determine its value.

Q.Five times a positive integer is 3 less than twice the square of that integer. Form the required quadratic equation to find the integer, and then solve the equation to determine its value.

Let the number be \(x\). According to the question: \(5x = 2x^2 - 3\) ⇒ \(2x^2 - 5x - 3 = 0\) ∴ The required quadratic equation is: **\(2x^2 - 5x - 3 = 0\)** Now solving: \(2x^2 - 5x - 3 = 0\) ⇒ \(2x^2 - 6x + x - 3 = 0\) ⇒ \(2x(x - 3) + 1(x - 3) = 0\) ⇒ \((x - 3)(2x + 1) = 0\) So, either \(x - 3 = 0\) ⇒ \(x = 3\) Or \(2x + 1 = 0\) ⇒ \(x = -\cfrac{1}{2}\) Since the number is a positive integer, The required number is \(x = 3\)