If five times a positive integer is 3 less than twice its square, then what is the value of that number?

Q.If five times a positive integer is 3 less than twice its square, then what is the value of that number?

Let the number be \(x\). According to the question: \[ 5x = 2x^2 - 3 \Rightarrow 2x^2 - 5x - 3 = 0 \Rightarrow 2x^2 - 6x + x - 3 = 0 \Rightarrow 2x(x - 3) + 1(x - 3) = 0 \Rightarrow (x - 3)(2x + 1) = 0 \] So, either \(x - 3 = 0 \Rightarrow x = 3\), or \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\) Since the number must be a positive integer, the number is 3.