Formula(/How) to find 2 numbers that add together to give one number and times to give another I have $2$ numbers $a, b$. I need a formula (or a how to) to find which $2$ numbers $c,d$ will add together to give a and times together to give $b$. 
So
$c + d = a$
$c \cdot d = b$ 
 A: Consider the quadratic
$$(x-c)(x-d) = x^2 -(c+d)x + cd = x^2 - ax +b$$
Thus the roots of $$x^2 -ax + b = 0$$ are your numbers $c$ and $d$.
The roots are given by
$$\frac{ a \pm \sqrt{a^2 -4b}}{2}$$
Check this page out for more information: Quadratic Equation.
A: Since I'm not so smart to imagine that the solution has anything to do with Vieta formulae or roots of polynomials,  ;-)  I'd try the following. First, let me call your unknowns $c,d$ with letters that look like unkowns (I've tried to do the problem with $c,d$ and at a certain point I forgot which among $a,b,c,d$ were the unkowns and which the data). So let $x = c$ and $y =d$. Your system becomes
$$
\begin{align}
x + y &= a \\
xy    &= b   \ .
\end{align}
$$
Now, do what you would do in those cases. For instance, solve the first equation for $x$ in terms of $y$:
$$
x = a -y  \ .
$$
Then substitute this expression for $x$ into the second equation:
$$
(a-y)y = b \ .
$$
And, surprise!, you've got a polynomial equation
$$
y^2 -ay + b = 0
$$
that looks like the ones I've been told. (So this Vieta must have been a really smart guy, but you've rediscovered his trick.)
And, of course, once you've got the solutions $y$ of this second degree equation, you should use $x = a-y$ to find $x$... (Just to find out that if $y$ is one of the roots of the last polynomial equation, $x$ is the other one.)
A: HINT $\rm\quad (X - c)\: (X - d) \; =\; X^2 - a\ X + b $
