t^2 + 1 = 13/6t?

Question

solve the equation

Answer 1

Assuming question is:-
t² + 1 = (13 / 6) t --------- ie the t is on top line.
6t² + 6 = 13 t
6t² - 13t + 6 = 0
(3t - 2) (2t - 3) = 0
t = 2 / 3 , t = 3 / 2

Answer 2

multiply both sides by 6 to eliminate the fraction
6*t^2 + 6 = 13*t
subtract 13t from both sides to set the right side to 0
6*t^2 + 6 - 13*t = 0
factor
(3*t - 2)*(2*t - 3) = 0
This makes 3*t - 2 = 0 or 2*t - 3 = 0
Thus, 3*t = 2 or 2*t = 3
t = 2/3 or t = 3/2 (which is 1.5)

Answer 3

t^2+1 =13/6t I assume 13/6t = (13/6)t?

t^2 -(13/6)t+1 = 0

= 13/2 +/- sqrt((13/6)^2 -4)/2
t = 0.167778, 1.8333