Question 309231
I think you meant to say rational equations. Radical equations look something like this {{{sqrt(x)=10}}}. Also, you can use parenthesis to show that 't-2' is underneath 1 by typing 1/(t-2)



{{{1/(t-2) = t/8}}} Start with the given equation.



{{{1*8 = t(t-2)}}} Cross multiply



{{{1*8 = t*t-t*2}}} Distribute



{{{8 = t^2-2t}}} Multiply



{{{0 = t^2-2t-8}}} Subtract 8 from both sides.



Notice that the quadratic {{{t^2-2t-8}}} is in the form of {{{At^2+Bt+C}}} where {{{A=1}}}, {{{B=-2}}}, and {{{C=-8}}}



Let's use the quadratic formula to solve for "t":



{{{t = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{t = (-(-2) +- sqrt( (-2)^2-4(1)(-8) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-2}}}, and {{{C=-8}}}



{{{t = (2 +- sqrt( (-2)^2-4(1)(-8) ))/(2(1))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{t = (2 +- sqrt( 4-4(1)(-8) ))/(2(1))}}} Square {{{-2}}} to get {{{4}}}. 



{{{t = (2 +- sqrt( 4--32 ))/(2(1))}}} Multiply {{{4(1)(-8)}}} to get {{{-32}}}



{{{t = (2 +- sqrt( 4+32 ))/(2(1))}}} Rewrite {{{sqrt(4--32)}}} as {{{sqrt(4+32)}}}



{{{t = (2 +- sqrt( 36 ))/(2(1))}}} Add {{{4}}} to {{{32}}} to get {{{36}}}



{{{t = (2 +- sqrt( 36 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{t = (2 +- 6)/(2)}}} Take the square root of {{{36}}} to get {{{6}}}. 



{{{t = (2 + 6)/(2)}}} or {{{t = (2 - 6)/(2)}}} Break up the expression. 



{{{t = (8)/(2)}}} or {{{t =  (-4)/(2)}}} Combine like terms. 



{{{t = 4}}} or {{{t = -2}}} Simplify. 



So the solutions are {{{t = 4}}} or {{{t = -2}}}