Answer
$[400,4400]$
Work Step by Step
$P=20x-2000-8x-0.0025x^2$,
$P=-0.0025x^2+12x-2000$,
for $P\geq2400$,
$-0.0025x^2+12x-2000\geq2400,$
$-0.0025x^2+12x-4400\geq0,$
Solving for the trinomial using quadratic formula for the equation of $ax^2+bx+c, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
thus, In this case $-0.0025x^2+12x-4400\geq0, x=\frac{-12\pm\sqrt{12^2-4(-0.0025)(-4400)}}{2(-0.0025)}=\frac{-12\pm10}{-0.005}$
Therefore, $x=4400$ or $x=400$.
$\begin{array}{lll}
Intervals&testing cases&-0.0025x^2+12x-4400&-0.0025x^2+12x-4400\geq0\\
(-\infty, 400) & 0& (-)&F\\
(400,4400) & 2000&(+)&T\\
(4400,\infty)&5000&(-)&F
\end{array}$
The Solution Set is:
$[400,4400]$
