What are the solutions to the equation t^2(5t^2-28t-9)=0 ?

The solutions to the equation t^2(5t^2-28t-9)=0 are t=0,t=(14+sqrt(241))/5 ,t=(14-sqrt(241))/5

Find the zeros of t^2(5t^2-28t-9)=0

The zeros of t^2(5t^2-28t-9)=0 are t=0,t=(14+sqrt(241))/5 ,t=(14-sqrt(241))/5

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t^{2} (5 t^{2} - 28 t - 9) = 0

t = 0, t = \frac{14 + 241}{5}, t = \frac{14 - 241}{5}

Decimal Notation

t = 0, t = 5.90483 \dots, t = - 0.30483 \dots

Solution steps

t^{2} (5 t^{2} - 28 t - 9) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

t = 0 or 5 t^{2} - 28 t - 9 = 0

Solve

5 t^{2} - 28 t - 9 = 0 : t = \frac{14 + 241}{5}, t = \frac{14 - 241}{5}

The solutions are

t = 0, t = \frac{14 + 241}{5}, t = \frac{14 - 241}{5}

x^{3} + 2 x^{2} - 21 x + 18 = 0

4 x^{4} - 29 x^{2} + 25 = 0

0 = x^{3} + 3 x^{2} - x - 3

- x^{3} - 2 x^{2} + 23 x + 60 = 0

5 = (6 x (x^{3} - 1)^{3})

What are the solutions to the equation t^2(5t^2-28t-9)=0 ?
The solutions to the equation t^2(5t^2-28t-9)=0 are t=0,t=(14+sqrt(241))/5 ,t=(14-sqrt(241))/5
Find the zeros of t^2(5t^2-28t-9)=0
The zeros of t^2(5t^2-28t-9)=0 are t=0,t=(14+sqrt(241))/5 ,t=(14-sqrt(241))/5