Answer
$16.797^\circ C$
Work Step by Step
Set $C$ to $150$ and solve for $x$ using the quadratic formula. Also disregard the temperature that is not in the given range of temperatures.
$\begin{align} 0.45x^2-1.65x+50.75&=150\\0.45x^2-1.65x-99.25&=0\\x&=\frac{-(-1.65\pm \sqrt{(-1.65)^2-4\cdot 0.45\cdot 50.75})}{2\cdot 0.45}\\&\approx 16.797\end{align}$
