Answer
$t=-2/3, 4/3$
Work Step by Step
$1+\frac{2}{3t-2}=\frac{8}{(3t-2)^2}$
$(3t-2)^2*1+(3t-2)^2*\frac{2}{3t-2}=(3t-2)^2*\frac{8}{(3t-2)^2}$
$(3t-2)^2+(3t-2)*2=8$
$3t*3t+3t(-2)+(-2)3t+(-2)(-2)+6t-4=8$
$9t^2-6t-6t+4+6t-4=8$
$9t^2-6t=8$
$9t^2-6t-8=8-8$
$9t^2-6t-8=0$
$t=(-b±\sqrt {b^2-4ac})/2a$
$t=(-(-6)±\sqrt {(-6)^2-4*9*(-8)})/2*9$
$t=(6±\sqrt {36+288)})/18$
$t=(6±\sqrt {324})/18$
$t=(6±18)/18$
$t=(6±18)/18$
$t=(6+18)/18$
$t=24/18$
$t=4/3$
$t=(6±18)/18$
$t=(6-18)/18$
$t=-12/18$
$t=-2/3$
