Answer
$t=-1+\displaystyle \frac{2\sqrt{3}}{3}$ or $t=-1-\displaystyle \frac{2\sqrt{3}}{3}$
Work Step by Step
$ 3t(t+2)=1\qquad$... use the distributive property: $a(b+c)=ab+ac$.
$3t\cdot t+3t\cdot 2=1$
$ 3t^{2}+6t=1\qquad$...add $-1$ to both sides.
$ 3t^{2}+6t-1=0\qquad$... solve with the Quadractic formula. $a=3,\ b=6,\ c=-1$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $6,\ a$ for $3$ and $c$ for $-1$.
$ t=\displaystyle \frac{-6\pm\sqrt{(6)^{2}-4\cdot(-1)\cdot 3}}{2\cdot 3}\qquad$... simplify.
$t=\displaystyle \frac{-6\pm\sqrt{36+12}}{6}$
$t=\displaystyle \frac{-6\pm\sqrt{48}}{6}$
$t=\displaystyle \frac{-6\pm\sqrt{4\cdot 4\cdot 3}}{6}$
$t=\displaystyle \frac{-6\pm 4\sqrt{3}}{6}$
$ t=-1\displaystyle \pm\frac{2\sqrt{3}}{3}\qquad$... the symbol $\pm$ indicates two solutions.
$t=-1+\displaystyle \frac{2\sqrt{3}}{3}$ or $t=-1-\displaystyle \frac{2\sqrt{3}}{3}$
