The fish population in a certain part of the ocean (in thousands of fish) as a function of the water's temperature (in degrees Celsius) is modeled by: \( p(x)=-2 x^{2}+40 x-72 \) Which temperatures will result in no fish (i.e. 0 population)? Enter the lower temperature first. Lower temperature: Higher temperature: \( \square \) degrees Celsius degrees Celsius

To find the temperatures at which the fish population is zero, we need to set the equation \( p(x) = -2x^2 + 40x - 72 \) to 0 and solve for \( x \): \[ -2x^2 + 40x - 72 = 0 \] Dividing the entire equation by -2 simplifies it to: \[ x^2 - 20x + 36 = 0 \] Now, using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -20 \), and \( c = 36 \): \[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} \] Calculating the discriminant: \[ (-20)^2 - 4 \cdot 1 \cdot 36 = 400 - 144 = 256 \] Now plug it back into the quadratic formula: \[ x = \frac{20 \pm \sqrt{256}}{2} \] \[ x = \frac{20 \pm 16}{2} \] This gives us two solutions: \[ x = \frac{36}{2} = 18 \quad \text{and} \quad x = \frac{4}{2} = 2 \] Thus, the temperatures that will result in no fish population are: Lower temperature: \( 2 \) degrees Celsius Higher temperature: \( 18 \) degrees Celsius