To determine how many sales Allison needs to make to earn [tex]$600, we need to solve the equation \( P(x) = 600 \), where \( P(x) \) is the net pay function given by \( P(x) = x^2 + 18x - 160 \).
1. Set up the equation:
\[
x^2 + 18x - 160 = 600
\]
2. Move 600 to the left side to set the equation to zero:
\[
x^2 + 18x - 160 - 600 = 0
\]
Simplifying the left side:
\[
x^2 + 18x - 760 = 0
\]
3. This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 18 \), and \( c = -760 \). To find the roots of this equation, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Plugging in the values of \( a \), \( b \), and \( c \):
\[
x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot (-760)}}{2 \cdot 1}
\]
Simplify inside the square root:
\[
x = \frac{-18 \pm \sqrt{324 + 3040}}{2}
\]
\[
x = \frac{-18 \pm \sqrt{3364}}{2}
\]
\[
x = \frac{-18 \pm 58}{2}
\]
4. Solve for the two possible values of \( x \):
\[
x_1 = \frac{-18 + 58}{2} = \frac{40}{2} = 20
\]
\[
x_2 = \frac{-18 - 58}{2} = \frac{-76}{2} = -38
\]
Since \( x \) represents the number of sales, it must be a non-negative number. Therefore, we discard the negative solution.
5. The number of sales Allison needs to make to earn \$[/tex]600 is:
[tex]\[
\boxed{20}
\][/tex]
