Answer
$t_{1}\approx-1.85$
$t_{2}\approx0.17$
Work Step by Step
The quadratic formula will be used, but before doing so substitution will be used:$1+t=u$ therefore:
$u^2\pi=\pi+u$
$\pi u^2-u-\pi=0$
$u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=\pi, b=-1, and\space c=-\pi$
$u=\frac{-(-1)\pm\sqrt{(-1)^2-4(\pi)(-\pi)}}{2(\pi)}$
$u=\frac{1\pm\sqrt{1+4\pi^2}}{2\pi}$
There are two solutions:
First:
$1+t=\frac{1+\sqrt{1+4\pi^2}}{2\pi}$
$t=\frac{1+\sqrt{1+4\pi^2}}{2\pi}-1$
$t\approx0.17$
Second:
$1+t=\frac{1-\sqrt{1+4\pi^2}}{2\pi}$
$t=\frac{1-\sqrt{1+4\pi^2}}{2\pi}-1$
$t\approx-1.85$
Plugging these values into the original equation does confirm that they are real solutions.