How to explain the minus (negative) sign before Kinetic energy operator in Hamiltonian operator? The Hamiltonian operator is normally written in this form:
\begin{align}
\large
H_\mathrm{operator}
= \ &
\large
\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}
&+ \quad &
\large V(x)
\\
&
\begin{array}{c}
\small\text{operator associated}\\
\small\text{with kinetic energy}
\end{array}
& &
\begin{array}{c}
\small\text{potential}\\
\small\text{energy}
\end{array}
\end{align}
How to explain the minus (negative) sign before Kinetic energy operator in Hamiltonian operator?
If I have to explain it theoretically, how do i explain it? The amount of energy required to give to an electron in order to give it enough kinetic energy to go out of the orbital? How KE value increases to compensate for PE when it goes down towards the nucleus?
 A: This Hamiltonian operator is derived from the classical one $H = \frac{p^2}{2m} + V(x)$ where $p$ and $x$ are replaced by the corresponding operators :
\begin{align}
\hat x \psi(x) &= x\psi(x) \\
\hat p \psi(x) &= -i\hbar \partial_x\psi(x)
\end{align}
The $-$ signs comes from the $i$ in the expression for $\hat p$, so your question boils down to :

Why is there an $i$ in the expression $\hat p = -i\hbar \partial_x$

@Photon gave one argument : you need it to give the right result on plane waves. @Jahan Claes gave another argument : you need it for $\hat p$ to be hermitian.
Looking at the kynetic energy $\hat T = - \frac{\hbar^2}{2m}\partial_x^2$ directly, you would expect it to be bounded below, ie for any wave function $\psi$ :
$$\langle \psi|\hat T|\psi\rangle = \int \text dx \psi^*(x)\hat T\psi(x)\geq 0$$
An integration by part shows this is the case only if you put in the $-$ sign.
A: You can see how the operator acts on a plain wave $\psi_k(x)=e^{ikx}$:
$$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} e^{ikx} = -\frac{\hbar^2}{2m}(ik)^2e^{ikx} = \frac{\hbar^2k^2}{2m}e^{ikx}$$
As you can see, the imaginary unit comes down twice when calculating the derivative and compensates the minus sign in the operator, so the eigenvalue is actually positive which suits our physical intuition: $E_k=\frac{\hbar^2k^2}{2m}$
A: If we have an operator whose representation in coordinate space is proportional to $\frac{\partial^2{}}{\partial{x}^2}$, and we need that the expectation value must be positive, we need the negative sign.
Indeed, integrating py parts:
$$
\int \psi^*(x)\frac{\partial^2{}}{\partial{x}^2}\psi(x) {\mathrm d}x=-\int \left(\frac{\partial{}}{\partial{x}}\psi^*(x)\right) \left(\frac{\partial{}}{\partial{x}}\psi(x)\right) {\mathrm d}x = -\int \left|\frac{\partial{}}{\partial{x}}\psi(x)\right|^2 {\mathrm d}x 
$$
A: I'm not sure I understand your specific question, but if you're wondering why there's a negative sign for the KE term which is conventionally positive, note that you have written down the Hamiltonian operator. In QM the Hamiltonian operator is used to  give you the total energy by acting on the wavefunction of the system (in this case). However it is not the total energy itself. The measurable quantity that we call energy is either an eigenvalue or an expectation value (depending on the system and the state of the system).
The two terms in the above Hamiltonian are the operators for the KE and the PE. They are not the energies themselves.
A: As @SolubleFish already said, the Hamiltonian operator is derived by $H = \frac{p^2}{2m} + V(x)$
Now, if you want to explain why $\hat p = -i\hbar \partial_x$, you could probably start by saying that you need the position and momentum operator to be canonical conjugates (this is important for the Heisenberg uncertainty principle and other things):
\begin{equation}
[\hat x, \hat p] = i\hbar
\end{equation}
Choosing the already suggested
\begin{align}
\hat x \psi(x) &= x\psi(x) \\
\hat p \psi(x) &= -i\hbar \partial_x\psi(x)
\end{align}
couple of operators, it can be easily shown that these two satisfy the precedent relation, but they aren't the only possible choice.
You can in fact choose the couple:
\begin{align}
\hat x' \psi(x) &= i\hbar \partial_p\psi(x) \\
\hat p' \psi(x) &= p\psi(x)
\end{align}
(where the definition of the momentum operator is more straightforward) and see that they satisfy $[\hat x', \hat p'] = i\hbar$.
Now, you can relate the $(\hat x, \hat p)$ and $(\hat x', \hat p')$ couple with a unitary transform (Stone-von Neumann theorem) which is, in this case, the Fourier transform: choosing the Fourier parameters to get $\hat x$ from $\mathscr{F}(\hat x')$, you can see that
\begin{equation}
\mathscr{F}[\hat p'\psi(p)](x) = \mathscr{F}[p\psi(p)](x) = -i\hbar \partial_x\mathscr{F}[\psi(p)](x)
\end{equation}
The way I see it, the presence of the minus sign in the KE term of the Hamiltonian operator written in the $x$ variable (or, equally, the presence of the imaginary unit in the definition of $\hat p$) is just a necessary condition to ensure that you can write the Hamiltonian in the $p$ variable as $\hat H\psi(p) = \dfrac{p^2}{2m}\psi(p) + V'(\hat p)\psi(p)$ without losing 'consistency'.
A: To be positive, has to be not the "operator", but the eigenvalues of such operator.
If you solve Schrodinger equation :
$-(h²/2m)d²/dx² f = Ef$
You find
$f=\exp i\sqrt k x$
With $k= 2mE/h²$
Hermitianity of p garantuees that only real E can be eigenvalue; the square root tells us that E has to be positive; furthermore, every E positiva has its eigenstate well definite.
