Question

Subtract seven more than twice a number from the square of one-third of the number you get zero

Answer 1

We will first translate this word problem into numerical expression.
Subtract seven more than twice a number from the square of one-third of the number and you get zero can be written into this form;
Let x = number
$(\frac{1}{3}) ^{2} - (2x + 7) = 0$
$\frac{1}{9} x^{2} - 2x - 7 = 0$
$9( \frac{1}{9} x^{2} -2x - 7) = 9 (0)$
$x^{2} - 18x - 63 = 0$
$(x - 21)(x + 3) = 0$

Therefore, the value of x are
 x - 21 = 0           and            x + 3 = 0
        x = 21                                  x = -3

Answer: {21, -3}