Answer
$t=-1+\displaystyle \frac{\sqrt{6}}{2}$ or $t=-1-\displaystyle \frac{\sqrt{6}}{2}$
Work Step by Step
$ 2t(t+2)=1\qquad$... use the distributive property: $a(b+c)=ab+ac$.
$2t\cdot t+2t\cdot 2=1$
$ 2t^{2}+4t=1\qquad$...add $-1$ to both sides.
$ 2t^{2}+4t-1=0\qquad$... solve with the Quadractic formula. $a=2,\ b=4,\ c=-1$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $4,\ a$ for $2$ and $c$ for $-1$.
$ t=\displaystyle \frac{-4\pm\sqrt{(4)^{2}-4\cdot(-1)\cdot 2}}{2\cdot 2}\qquad$... simplify.
$t=\displaystyle \frac{-4\pm\sqrt{16+8}}{4}$
$t=\displaystyle \frac{-4\pm\sqrt{24}}{4}$
$t=\displaystyle \frac{-4\pm\sqrt{4\cdot 6}}{4}$
$t=\displaystyle \frac{-4\pm 2\sqrt{6}}{4}$
$ t=-1\displaystyle \pm\frac{\sqrt{6}}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$t=-1+\displaystyle \frac{\sqrt{6}}{2}$ or $t=-1-\displaystyle \frac{\sqrt{6}}{2}$
