This is a good example of a question with answers of very different levels of mathematical sophistication! Since you say nothing about this, let me try an elementary one.
What you call the Dirac delta function (which is not a function, at least not in the sense of a function from $\mathbb R$ to $\mathbb R$) is a strange object but something about it is clear:
One asks that $\displaystyle\int_y^z\delta(x)\mathrm dx=0$ if $y\leqslant z<0$ or if $0<y\leqslant z$ and that $\displaystyle\int_y^z\delta(x)\mathrm dx=1$ is $y<0<z$.
We will not use anything else about the Dirac $\delta$.
If one also asks that $\displaystyle\int_y^zu''(x)\mathrm dx=u'(z)-u'(y)$ for every $y\leqslant z$, one can integrate once your equation $\color{red}{-u''=\delta}$, getting
that there exists $a$ such that
$$
u'(x)=a-[x\geqslant0],
$$
where we used Iverson bracket notation. Now let us integrate this once again.
Using the facts that $\displaystyle\int_y^zu'(x)\mathrm dx$ should be $u(z)-u(y)$ for every $y\leqslant z$, and the value of $\displaystyle\int_y^z[x\geqslant0]\mathrm dx$, one gets that for every fixed negative number $x_0$,
$$
u(x)=u(x_0)+a(x-x_0)-x\cdot[x\geqslant0].
$$
This means that $b=u(x_0)-ax_0$ does not depend on $x_0<0$, hence finally, for every $x$ in $\mathbb R$,
$$
\color{red}{u(x)=ax+b-x\cdot[x\geqslant0]}.
$$
(And the condition that $u(-2)=u(3)=0$ imposes that $a=3/5$ and $b=6/5$.)
This is the general solution of the equation $-u''=\delta$. Note that every solution $u$ is $C^\infty$ on $\mathbb R\setminus\{0\}$ but only $C^0$ at $0$ hence $u'$ and $u''$ do not exist in the rigorous sense usually meant in mathematics. Note finally that $u$ is also
$$
u(x)=ax+b-x\cdot[x\gt0].
$$
