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}
