Question

Nine times a certain whole number is equal to five less than twice the
square of the number. Then the number is​

Answer 1

Answer:

$Let \: the \: number \: be \: x$

$9x = 2 {x}^{2} - 5$

$2 {x}^{2} - 9x - 5 = 0$

$2 {x}^{2} + x - 10x - 5 = 0$

$x(2x + 1) - 5(2x + 1) = 0$

$(2x + 1)(x - 5) = 0$

$2x + 1 = 0$

$x = - \frac{1}{2}$

$x - 5 = 0$

$x = 5$

$So \: the \: number \: can \: be \:both \: 5 \: and \: - \frac{1}{2} .$