Find the real solutions, if any, -4.9*t^5 + 1.90*t^4 + 4.80*t^3?

Answer 1

-4.9t^5 + 1.90t^4 + 4.80t^3
= -t^3 (4.9t^2 - 1.90t - 4.80)
= 0 <=> t = 0 or 4.9t^2 - 1.90t - 4.80 = 0
<=> t = 0 or (1.90 ± √(1.90^2 - 4(4.9)(-4.80))) / (9.8)
<=> t = 0, 1.20, -0.81 (2 d.p.)

Answer 2

For 'solutions' you need to have an equation, so I will assume you mean
-4.9*t^5 + 1.90*t^4 + 4.80*t^3 = 0

Get rid of the awkward negative at the start (multiply through by -1):
4.9*t^5 - 1.90*t^4 - 4.80*t^3 = 0

Factorise by taking out t^3
t^3 (4.9*t^2 - 1.90*t - 4.80) = 0

Clearly t=0 is a solution (since it makes t^3 equal to 0).
The other solutions will come when the term in the bracket is 0. So you just have to solve the quadratic
4.9*t^2 - 1.90*t - 4.80 = 0.

Answer 3

I am going to assume that this is equal to 0.

Factor out a t^3

t^3(-4.9t^2 + 1.9t + 4.8)

So t = 0 is one solution

t = (-1.9 +/- SQRT((1.9)^2 - 4(4.8)(-4.9)))/2(-4.9)
t = (-1.9 +/- SQRT(97.69))/(-4.9)

t = 0.3877 +/- 4.0687
t = 4.4564 and t = -3.681 are two others