Answer
$r_{1}\approx-1.44$
$r_{2}\approx0.44$
Work Step by Step
The quadratic formula will be used, but before doing so substitution will be used:$1+r=u$ therefore:
$\pi u^2=2+\pi u$
$\pi u^2-\pi u-2=0$
$u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=\pi, b=-\pi, and\space c=-2$
$u=\frac{-(-\pi)\pm\sqrt{(-\pi)^2-4(\pi)(-2)}}{2(\pi)}$
$u=\frac{\pi\pm\sqrt{\pi^2+8\pi}}{2\pi}$
There are two solutions:
First:
$1+r=\frac{\pi+\sqrt{\pi^2+8\pi}}{2\pi}$
$r=\frac{\pi+\sqrt{\pi^2+8\pi}}{2\pi}-1$
$r\approx0.44$
Second:
$1+r=\frac{\pi-\sqrt{\pi^2+8\pi}}{2\pi}$
$r=\frac{\pi-\sqrt{\pi^2+8\pi}}{2\pi}-1$
$r\approx-1.44$
Plugging these values into the original equation with the calculator does confirm that they are real solutions.