Newtons Law of Cooling: In a Murder Investigation, the Temperature of the Corpse was 32.5 Degrees Celsius at 1.30 PM and 30.3 Degree Celsius In a Murder Investigation, the temperature of the corpse was $32.5^\circ C$ at $1:30$ PM and $30.3^\circ C$ an hour later. Normal body temperature is $37^\circ C$ and the temperature of the surroundings was $20^\circ C$. When did the murder take place? 
Newton's Law of Cooling $$k(T-T_s)$$
$$k(37-20)$$
$$k=17$$
Where do I go from here?
 A: You made several mistakes already. We can set up this problem like this:
$$\frac{dT}{dt}=-k(T-T_s)=-k(T-20),T(0)=32.5,T(1)=30.3$$
Here $0$ corresponds to the first measurement and $1$ corresponds to the second measurement an hour later*. You need to solve this equation (which will take some calculus, not just algebra as you did), and then use the two given temperatures to compute $k$. Once you know $k$, then you fully know $T(t)$, and you can solve the equation $T(t)=37$ for $t$.
* There are other ways to choose the time coordinates; for instance, you could make $0$ be the time of the murder if you wanted. This slightly changes the rest of the problem.
A: Newton's law of cooling:
$$
\frac{dT}{dt} = -k(T-T_s).
$$
Solution:
$$
T = T_s + (T_0 - T_s)e^{-kt},
$$
where $T_0 = T(0)$. In your case, $T_0=32.5$, $T_s = 20$. So,
$$
T = 20 + 12.5e^{-kt}.
$$
At $t=1$ (hour) must be $T = 30.3$. Therefore
$$
30.3 = 20 + 12.5e^{-k}\Longrightarrow k = -\ln\Big(\frac{103}{125}\Big) \approx 0.1935847491
$$
If $T=37$ at $t=t_m$, then
$$
37 = 20 + 12.5e^{-kt_m}\Longrightarrow t_m = -\frac1k\ln\Big(\frac{34}{25}\Big) = \frac{\ln\big(\frac{34}{25}\big)}{\ln\big(\frac{103}{125}\big)}\approx -1.58837254
$$
In minutes, $t_m \approx -95$. I think, you can proceed ;)
A: Notice, according to newton's law of cooling, rate of cooling $\frac{dT}{dt}$ of a body is directly proportional to the temperature difference $(T-T_{\infty})$ between system & surrounding 
$$\frac{dT}{dt}=-k(T-T_{\infty})$$ 
Where, $T$ is the instantaneous temperature & $T_{\infty}$ is surrounding temperature. The negative sign indicates the decrease in the temperature w.r.t. time $t$. Now we have
$$\frac{dT}{T-T_{\infty}}=-kdt$$ $$\int\frac{dT}{T-T_{\infty}}=-\int kdt$$ 
$$\ln(T-T_{\infty})=-kt+C$$
Now, at time $t=0$ the initial temperature is $T_i$ then we have $$\ln(T_i-T_{\infty})=-k(0)+C\iff C=\ln(T_i-T_{\infty})$$ Hence, we have $$\ln(T-T_{\infty})=-kt+\ln(T_i-T_{\infty})$$
$$\ln\left(\frac{T-T_{\infty}}{T_i-T_{\infty}}\right)=-kt$$
$$\frac{T-T_{\infty}}{T_i-T_{\infty}}=e^{-kt}$$
Condition 1: Temperature fall $\color{blue}{32.5^\circ\ C \to 30.3^\circ\ C}$ 
Setting the values, $T=30.3$, $T_{i}=32.5$, $T_{\infty}=20$ & $t=1\ hr$ 
$$\frac{30.3-20}{32.5-20}=e^{-k(1)}$$ $$e^{-k}=\frac{103}{125}\iff k=\ln\left(\frac{125}{103}\right)$$
Condition 2: Temperature fall $\color{blue}{37^\circ\ C \to 32.5^\circ\ C}$ 
Setting the values, $T=32.5$, $T_{i}=37$, $T_{\infty}=20$, we get
$$\frac{32.5-20}{37-20}=e^{-kt}$$ $$e^{-kt}=\frac{25}{34}\iff t=\frac{1}{k}\ln\left(\frac{34}{25}\right)$$ Now, setting the value of $k$, we get $$t=\frac{\ln\left(\frac{34}{25}\right)}{\ln\left(\frac{125}{103}\right)}=1.5884\ hrs$$ 
Since, the murder took place at body temperature $37^\circ\ C$ & becomes $32.5^\circ\ C$ at $1:30$ PM after  $1.5884\ hrs=95.3023\ minutes$
Hence, we conclude that 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{murder took place around}\ 11:54:42\ PM}}$$
