What are the solutions to the equation 10.4=19t-4.9t^2 ?

The solutions to the equation 10.4=19t-4.9t^2 are t=(95-sqrt(3929))/(49),t=(95+sqrt(3929))/(49)

Find the zeros of 10.4=19t-4.9t^2

The zeros of 10.4=19t-4.9t^2 are t=(95-sqrt(3929))/(49),t=(95+sqrt(3929))/(49)

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10.4 = 19 t - 4.9 t^{2}

t = \frac{95 - 3929}{49}, t = \frac{95 + 3929}{49}

Decimal Notation

t = 0.65955 \dots, t = 3.21799 \dots

Solution steps

10.4 = 19 t - 4.9 t^{2}

Multiply both sides by

10

104 = 190 t - 49 t^{2}

Switch sides

190 t - 49 t^{2} = 104

Move

104

to the left side

190 t - 49 t^{2} - 104 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 49 t^{2} + 190 t - 104 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 190 \pm 19 0 ^{2} - 4 ( - 49 ) ( - 104 )}{2 ( - 49 )}

19 0^{2} - 4 (- 49) (- 104) = 23929

t_{1, 2} = \frac{- 190 \pm 2 3929}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 190 + 2 3929}{2 ( - 49 )}, t_{2} = \frac{- 190 - 2 3929}{2 ( - 49 )}

t = \frac{- 190 + 2 3929}{2 ( - 49 )} : \frac{95 - 3929}{49}

t = \frac{- 190 - 2 3929}{2 ( - 49 )} : \frac{95 + 3929}{49}

The solutions to the quadratic equation are:

t = \frac{95 - 3929}{49}, t = \frac{95 + 3929}{49}

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pq - 3 r + p r - 3 q

5 0^{2} + 12 0^{2} = c^{2}

What are the solutions to the equation 10.4=19t-4.9t^2 ?
The solutions to the equation 10.4=19t-4.9t^2 are t=(95-sqrt(3929))/(49),t=(95+sqrt(3929))/(49)
Find the zeros of 10.4=19t-4.9t^2
The zeros of 10.4=19t-4.9t^2 are t=(95-sqrt(3929))/(49),t=(95+sqrt(3929))/(49)