Equation with two unknowns I have this equation to solve and I solved it but don't know if the result is correct.
$\begin{cases}2\pi r_1+2\pi r_2=24 & (1) \\ \pi r_1^2+\pi r_2^2=20 & (2)\end{cases}$
Equation $(1)$ gives $r_1=\frac{12-\pi r_2}{\pi}$
Plug that into $(2)$, we have $\pi\big(\frac{12-\pi r_2}{\pi}\big)^2+\pi r_2^2=20 \Rightarrow 2\pi r_2^2-24r_2+\frac{144}{\pi}-20=0$.
$\frac {2\pi r_2^2-24r_2+\frac{144}{\pi}-20}{2\pi}=0$
So to solve the unknowns I use the reduced quadratic equation
$r_2^2 - 3.819r_2 + 4.11 = 0$
$r = - \frac {- 3.819}{2} {+ \choose -} \sqrt{\frac {- 3.819^2}{2^2}-4.11}$
$r = 1.9 {+ \choose -} \sqrt{-0.463}$
So because I don't have complex numbers in my course I think that this equation does not have an solution.
I'm I correct in my assumption or is my assumption wrong?
Thanks!!

Let's consider the system
\begin{cases}
x+y=a\\
x^2+y^2=b
\end{cases}
which is the same as yours with $a=12/\pi$ and $b=20/\pi$. We can rewrite the second equation as
$$
(x+y)^2-2xy=b
$$
so, taking into account the first equation, it becomes
$$
2xy=a^2-b
$$
Now the problem is reduced to finding two numbers of which we know the sum and the product: they are the roots of
$$
z^2-az+\frac{a^2-b}{2}=0
$$
The discriminant is
$$
a^2-4\frac{a^2-b}{2}=a^2-2a^2+2b=2b-a^2
$$
so we have real solutions if and only if
$$
2b-a^2\ge0
$$
In your case
$$
2\frac{20}{\pi}-\frac{144}{\pi^2}=\frac{40\pi-144}{\pi^2}
$$
and $144/40=3.6>\pi$, so the discriminant is negative.
