what would be the way to solve a system of equations like this one? Solve:
$xy=-30$
$x+y=13$
{15, -2} is a particular solution, but, how would I know if is the only solution, or what would be the way to solve this without "guessing" ?
 A: $$xy=-30$$
$$x+y=13$$
from second equation $y=13-x$ then put on first we get
$$x(13-x)=-30$$
$$x^2-13x-30=0$$
$$x_{1,2}=\frac{13\pm\sqrt{189}}{2}=\frac{13\pm17}{2}$$
$$x_1=\frac{13+17}{2}=15,x_2=\frac{13-17}{2}=-2$$
$$y_1=13-15=-2,y_2=13-(-2)=15$$
so the solutions are $(15,-2)$ and $(-2,15)$
A: The other answers here show the strategies of:

$1$) Substitution (the most general). See Kim Jong Un's answer on this question.

$2)$ Solving a quadratic equation that has the roots $x,y$ (this uses the Vieta's formulas). See Sami Ben Romdhane's answer.

$3)$ Using the fact that $(x+y)^2-4xy=(x-y)^2$ to have the numeric value of $x-y$. Then $(x+y)-(x-y)=2y$, an easy way to get the value of $y$. See André Nicolas' answer.

I'm posting another method that works great for this particular problem.
Add the equations in a specific way to get:$$\begin{align}xy+2(x+y)&=-30+2\cdot 13=-4\\\iff (x+2)(y+2)&=0\end{align}$$
Thus either $x=-2$ or $y=-2$, which give the solutions $(-2,15)$ and $(15,-2)$, respectively.
A: We have in general
$$(x+y)^2-4xy=(x-y)^2.$$
In our case that gives $(x-y)^2=289$, so $x-y=\pm 17$.
Now solve the system $x+y=13$, $x-y=17$ and the system $x+y=13$, $x-y=-17$.
Remark: This procedure for finding $x$ and $y$ given their sum and product goes back to Neo-Babylonian times. (Of course, algebraic notation was not used, but an equivalent algorithm was taught.)
A: Start with
$$
x+y=13 \iff y=13-x
$$ 
so that
$$
-30=xy=x(13-x)=13x-x^2 \iff x^2-13x-30=0
$$
which solves to give $x=-2$ and $x=15$. The corresponding $y$ values are $15$ and $-2$.
A: The solutions of the system of equations $xy=p$ and $x+y=s$ are the roots of the quadratic equation:
$$x^2-sx+p=0$$
so in your case we solve
$$x^2-13x-30=0$$
using the discriminant:
$$\Delta=13^2+4\times 30=17^2$$ 
so the two roots are
$$x_1=\frac{13+17}{2}=15\quad;\quad x_2=\frac{13-17}{2}=-2$$
A: The sum
$$x + y = 13 = -a$$
and the product
$$xy = -30 = b$$
gives rise to the quadratic equation
$$z^2 + az + b = 0.$$
Consequently, we have to solve the quadratic equation
$$z^2 - 13z - 30 = 0$$
for $z$, in order to get $x$ and $y$.
Since the discriminant $a^2 - 4b = {17}^2$, $z^2 - 13z - 30$ is factorable as:
$$z^2 - 13z - 30 = (z - 15)(z + 2).$$
Setting it equal to zero, we get the roots: $z = 15$ or $z = -2$.
Since the equations
$$x + y = 13$$
and
$$xy = -30$$
are symmetric in $x$ and $y$, we get the solutions:
$$x = 15, y = -2$$
or
$$x = -2, y = 15.$$
