Ive the equation by using substitution. Express numbers in simplified form using integers, fractions, or radicals.
3t^2/5+ 7t^1/5+2 = 0
Let's substitute u = t^(1/5):
3u^2 + 7u + 2 = 0
Now we can factor this quadratic equation:
(3u + 1)(u + 2) = 0
Setting each factor to zero:
3u + 1 = 0
u = -1/3
u + 2 = 0
u = -2
Now substituting back in u = t^(1/5):
t^(1/5) = -1/3
t = (-1/3)^5
t = -1/243
t^(1/5) = -2
t = (-2)^5
t = -32
Therefore, the solutions to the equation 3t^2/5 + 7t^1/5 + 2 = 0 are t = -1/243 and t = -32.