## Essential University Physics: Volume 1 (3rd Edition)

$3.97^{\circ}C$
We simplify the equation to find T using the quadratic formula and obtain: $T = \frac{-b \pm \sqrt{b^2-4ca}}{2c}$ Using the values for a, b, and c, we find: $T = \frac{-1.7 \times 10^{-5} \pm \sqrt{(-1.7 \times 10^{-5})^2-4(-6.43 \times 10^{-5})(-2.02\times 10^{-7})}}{2(-2.02\times 10^{-7})}=$$3.97^{\circ}C$