EDIT
Recall the statement of the intermediate value theorem.
Theorem If $f(x)$ is a real-valued continuous function on the interval $[a,b]$, then given any $y \in [\min(f(a),f(b)), \max(f(a),f(b))]$, there exists $c \in [a,b]$ such that $f(c) = y$.
The theorem guarantees us that given any value $y$ in-between $f(a)$ and $f(b)$, the continuous function $f(x)$ takes the value $y$ for some point in the interval $[a,b]$.
Now lets get back to our problem. Look at the function $f(x) = \cos(x) - x$.
We have $f(0) = 1 > 0$.
We also have that $f(1) = \cos(1) - 1$. But $\cos(x) < 1$, $\forall x \neq2 n\pi$, where $n \in \mathbb{Z}$. Clearly, $1 \neq 2 n \pi$, where $n \in \mathbb{Z}$. Hence, we have that $\cos(1) < 1 \implies f(1) < 0$.
Hence, we have a continuous function $f(x) = \cos(x) - x$ on the interval $[0,1]$ with $f(0) = 1$ and $f(1) = \cos(1) - 1<0$. ($a=0$, $b=1$, $f(a) = 1$ and $f(b) = \cos(1) -1 < 0$).
Note that $0$ lies in the interval $[\cos(1)-1,1]$. Hence, from the intermediate value theorem, there exists a $c \in [0,1]$ such that $f(c) = 0$.
This means that $c$ is a root of the equation. Hence, we have proved that there exists a root in the interval $[0,1]$.
I don´t know what you mean by $ Q$. I will use the numbers which are mentioned the website you´ve posted. But first of all I solve the general formula for the time ($n$).
$$L\cdot\left( \left( 1+\frac{i}{m}\right)^{m}\right)^n=p\cdot \frac{\left(\left(1+\frac{i}{m}\right)^{m}\right)^n-1}{\frac{i}{m}}$$
$L$ is the loan amount. $p$ is the regular payment. $i$ is the yearly interest rate. $n$ is the number of years.
$m$ divides the year in $m$ equal parts. If the payment is made and compunded $\color{orange}{\textrm{monthly}}$ ($\color{blue}{\textrm{daily}}$, $\color{green}{\textrm{semi-annually}}$, $\color{red}{\textrm{yearly}}$) then $m$ is $\color{orange}{\textrm{12}}$ ($\color{blue}{\textrm{365}}$, $\color{green}{\textrm{2}}$, $\color{red}{\textrm{1}}$)
Dividing the equation by $p$ and multiplying it by $\frac{i}{m}$
$$\frac{L}{p}\cdot \frac{i}{m}\cdot \left( \left( 1+\frac{i}{m}\right)^{m}\right)^n= \left(\left(1+\frac{i}{m}\right)^{m}\right)^n-1$$
Adding $1$ and subtracting the whole left side.
$$1= \left(\left(1+\frac{i}{m}\right)^{m}\right)^n-\frac{L}{p}\cdot \frac{i}{m}\cdot \left( \left( 1+\frac{i}{m}\right)^{m}\right)^n$$
Factoring out the term in the brackets.
$$1= \left(\left(1+\frac{i}{m}\right)^{m}\right)^n\cdot \left( 1-\frac{L}{p}\cdot \frac{i}{m}\right)$$
$$\frac1{ 1-\frac{L}{p}\cdot \frac{i}{m}}= \left(\left(1+\frac{i}{m}\right)^{m}\right)^n $$
Taking logs. I use the logarithm naturalis here, which is based on $e$. But you can take other logarithms as well.
$$\ln\left(\frac1{ 1-\frac{L}{p}\cdot \frac{i}{m}}\right)=\ln\left(\left(\left(1+\frac{i}{m}\right)^{m}\right)^n\right)$$
The exponent n can be put infront of the $\ln$-function, due logarithm rule $\ln(a^n)=n\cdot \ln(a)$
$$\ln\left(\frac1{ 1-\frac{L}{p}\cdot \frac{i}{m}}\right)=n\cdot \ln\left(\left(1+\frac{i}{m}\right)^{m}\right)$$
For the left side we can use that $\ln\left(\frac1a\right)=-\ln(a)$
$$-\ln\left( 1-\frac{L}{p}\cdot \frac{i}{m}\right)=n\cdot \ln\left(\left(1+\frac{i}{m}\right)^{m}\right)$$
$$n=\frac{-\ln\left( 1-\frac{L}{p}\cdot \frac{i}{m}\right)}{\ln\left(\left(1+\frac{i}{m}\right)^{m}\right)}=-\frac{\ln\left( 1-\frac{L}{p}\cdot \frac{i}{m}\right)}{m\cdot\ln\left(1+\frac{i}{m}\right)}$$
The given numbers of the linked site $L=28400, p=297, m=\color{orange}{\textrm{12}}$ and $i=0.0466$ we obtain
$$n=-\frac{\ln\left( 1-\frac{28400}{297}\cdot \frac{0.0466}{12}\right)}{12\cdot\ln\left(1+\frac{0.0466}{12}\right)}=9.97979 \approx 10$$
Remark
At the calculator the payment of $p=297$ has been rounded to dollars, without cents. I have calculated with the formula above a payment of $296.528$. See here the equation and the result. If I use this value of $p$ the result for $n$ is even more closer to $10$. Nevertheless $n$ must be an integer.
Best Answer
$$f(x)=x^5−x^2+2x+3$$
As you can see $f(0)=3>0$ and $f(-1)=-1<0$
Thus there is at least one root of $f(x)=0 $ in Interval $(-1,0)$
Now calculate the value of $$f(-\frac{1}{2})=\frac{55}{32}>0$$
Thus now our interval is shortened and it is $(-1,-\frac{1}{2})$
$$f(-\frac{3}{4})=\frac{717}{1024}>0$$
Our interval is now $(-1,-\frac{3}{4})$
$$f(-\frac{7}{8})=-\frac{935}{32768}<0$$
Our interval is now $(-\frac{7}{8},-\frac{3}{4})$
similarly, keep doing until you get the desired result