What are the solutions to the equation -(9.81)/2 t^2+216.5t-45.5t=0 ?

The solutions to the equation -(9.81)/2 t^2+216.5t-45.5t=0 are t=0,t= 3800/109

Find the zeros of -(9.81)/2 t^2+216.5t-45.5t=0

The zeros of -(9.81)/2 t^2+216.5t-45.5t=0 are t=0,t= 3800/109

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- \frac{9.81}{2} t^{2} + 216.5 t - 45.5 t = 0

t = 0, t = \frac{3800}{109}

Decimal Notation

t = 0, t = 34.86238 \dots

Solution steps

- \frac{9.81}{2} t^{2} + 216.5 t - 45.5 t = 0

Multiply both sides by

100

- \frac{981}{2} t^{2} + 17100 t = 0

Multiply both sides by

2

- 981 t^{2} + 34200 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 34200 \pm 3420 0 ^{2} - 4 ( - 981 ) \cdot 0}{2 ( - 981 )}

3420 0^{2} - 4 (- 981) \cdot 0 = 34200

t_{1, 2} = \frac{- 34200 \pm 34200}{2 ( - 981 )}

Separate the solutions

t_{1} = \frac{- 34200 + 34200}{2 ( - 981 )}, t_{2} = \frac{- 34200 - 34200}{2 ( - 981 )}

t = \frac{- 34200 + 34200}{2 ( - 981 )} : 0

t = \frac{- 34200 - 34200}{2 ( - 981 )} : \frac{3800}{109}

The solutions to the quadratic equation are:

t = 0, t = \frac{3800}{109}

12 5^{\frac{5}{3}}

- \frac{1}{6} \cdot 2

(9 a b^{2} c^{4}) (- 11 a^{3} b) (- 2 b^{2} c^{5})

\frac{5 a ^{2} b ^{3} c ^{4}}{7 a b ^{3} c ^{6}}

- x \leq 14

What are the solutions to the equation -(9.81)/2 t^2+216.5t-45.5t=0 ?
The solutions to the equation -(9.81)/2 t^2+216.5t-45.5t=0 are t=0,t= 3800/109
Find the zeros of -(9.81)/2 t^2+216.5t-45.5t=0
The zeros of -(9.81)/2 t^2+216.5t-45.5t=0 are t=0,t= 3800/109