I know this might sound dumb but I can't find $t$ for this equation :
$ -288.45e^{-t/4.5} -44.1t + 318.45=0 $
I tried using the common way of solving exponential (with $ln$) but can't seem to factorize the $t$'s together. Any help is greatly appreciated!
On
Let us begin by writing your equation as follows: $$-ae^{-2t/9}-bt + c = 0, \tag1$$ where $a = 288.45, \quad b = 44.1, \quad c = 318.45$.
Now let me outline how one would solve this equation in terms of the Lambert W function. The idea behind this method is we need to rewrite (1) so that it is in the form for the defining equation for the Lambert W function $\text{W}(x)$, namely $$\text{W}(x) e^{\text{W}(x)} = x. \tag2$$
Doing so we have \begin{align*} -a e^{-2t/9} &= bt - c\\ e^{2t/9} (bt - c) &= -a\\ e^{-2t/9} \left (\frac{2t}{9} - \frac{2c}{9b} \right ) &= -\frac{2a}{9b}\\ \exp \left (\frac{2t}{9} - \frac{2c}{9b} \right ) \left (\frac{2t}{9} - \frac{2c}{9b} \right ) &= -\frac{2a}{9b} \exp \left (-\frac{2c}{9b} \right ). \end{align*} As the above equation is now exactly in the form of (2), its solution in terms of the Lambert W function will be $$\frac{2t}{9} - \frac{2c}{9b} = \text{W}_\nu \left (-\frac{2a}{9b} \exp \left [-\frac{2c}{9b} \right ] \right ),$$ or $$t = \frac{c}{b} + \frac{9}{2} \text{W}_\nu \left (-\frac{2a}{9b} \exp \left [-\frac{2c}{9b} \right ] \right ).$$ Here $\nu$ denotes the various branches for the Lambert W function. In your case, as the argument to the Lambert W function lies between $-1/e$ and zero, for real solutions the two real branches of $\nu = 0$ (the principal branch) and $\nu = -1$ (the secondary real branch) are selected. Thus $$t = \begin{cases} \dfrac{c}{b} + \dfrac{9}{2} \text{W}_0 \left (-\dfrac{2a}{9b} \exp \left [-\dfrac{2c}{9b} \right ] \right )\\[2ex] \dfrac{c}{b} + \dfrac{9}{2} \text{W}_{-1} \left (-\dfrac{2a}{9b} \exp \left [-\dfrac{2c}{9b} \right ] \right ) \end{cases} $$
With your particular parameters, numerically one has $$t = \begin{cases} 5.12852532\ldots\\ -1.06351637\ldots \end{cases}$$
Let us suppose that you do not want to use Lambert function.
So, consider that you look for the zero(s) of function $$f(t)=-288.45e^{-t/4.5} -44.1t + 318.45$$ To make it nicer, let us multiply all terms by $-20$ to get $$f(t)=5769 \,e^{-2 t/9}+882 t-6369$$ Computing the derivatives $$f'(t)=882-1282 e^{-2 t/9}$$ $$f''(t)=\frac{2564}{9} e^{-2 t/9}\qquad > 0 \qquad \forall t$$
The first derivative cancels for $$t_*=\frac{9}{2} \log \left(\frac{641}{441}\right)\approx1.68293\implies f(t_*)=3969 \log \left(\frac{641}{441}\right)-2400\approx -915.655$$ So, you have two roots for the equation such that $t_1 < t_*$ and $t_2 >t_*$.
Just to get estimates of the roots, let us perform a Taylor expansion around $t_*$. Since $f'(t_*)=0$, we shall have $$f(t)=f(t_*)+\frac12 f''(t_*)(t-t_*)^2+O\left( (t-t_*)^3\right)$$ Solving for $0$, then the estimates are $$\frac{1}{14} \left(63 \log \left(\frac{641}{441}\right)-\sqrt{6 \left(800-1323 \log \left(\frac{641}{441}\right)\right)}\right)\approx -1.37377 $$ $$\frac{1}{14} \left(63 \log \left(\frac{641}{441}\right)+\sqrt{6 \left(800-1323 \log \left(\frac{641}{441}\right)\right)}\right)\approx 4.73963 $$
So, we are ready for Newton method. Applied to your case, the iterates will then be given by $$t_{n+1}=\frac{6369\, e^{2 t_n/9}-641 (2 t_n+9)}{882 \,e^{2 t_n/9}-1282}$$
So, for the first root, we shall get as iterates $$\left( \begin{array}{cc} n & t_n \\ 0 & -1.373770280 \\ 1 & -1.084720057 \\ 2 & -1.063624948 \\ 3 & -1.063516376 \\ 4 & -1.063516374 \end{array} \right)$$ which is the solution for ten significant figures.
For the second root, the iterates will be $$\left( \begin{array}{cc} n & t_n \\ 0 & 4.739631513 \\ 1 & 5.145318766 \\ 2 & 5.128552404 \\ 3 & 5.128525323 \end{array} \right)$$ which is the solution for ten significant figures.