What are the solutions to the equation 30=67t-5t^2 ?

The solutions to the equation 30=67t-5t^2 are t=(67-sqrt(3889))/(10),t=(67+sqrt(3889))/(10)

Find the zeros of 30=67t-5t^2

The zeros of 30=67t-5t^2 are t=(67-sqrt(3889))/(10),t=(67+sqrt(3889))/(10)

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30 = 67 t - 5 t^{2}

t = \frac{67 - 3889}{10}, t = \frac{67 + 3889}{10}

Decimal Notation

t = 0.46381 \dots, t = 12.93618 \dots

Solution steps

30 = 67 t - 5 t^{2}

Switch sides

67 t - 5 t^{2} = 30

Move

30

to the left side

67 t - 5 t^{2} - 30 = 0

Write in the standard form

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

- 5 t^{2} + 67 t - 30 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 67 \pm 6 7 ^{2} - 4 ( - 5 ) ( - 30 )}{2 ( - 5 )}

6 7^{2} - 4 (- 5) (- 30) = 3889

t_{1, 2} = \frac{- 67 \pm 3889}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 67 + 3889}{2 ( - 5 )}, t_{2} = \frac{- 67 - 3889}{2 ( - 5 )}

t = \frac{- 67 + 3889}{2 ( - 5 )} : \frac{67 - 3889}{10}

t = \frac{- 67 - 3889}{2 ( - 5 )} : \frac{67 + 3889}{10}

The solutions to the quadratic equation are:

t = \frac{67 - 3889}{10}, t = \frac{67 + 3889}{10}

7 a + 9 = 9

9 x^{2} - 42 x y + 49 y^{2}

(\frac{1}{8})^{- \frac{4}{3}}

(- 9 n^{4} + 8) - (- 9 n^{4} + 2 n^{3} + 5 n^{2})

0.05 \cdot 9.8

What are the solutions to the equation 30=67t-5t^2 ?
The solutions to the equation 30=67t-5t^2 are t=(67-sqrt(3889))/(10),t=(67+sqrt(3889))/(10)
Find the zeros of 30=67t-5t^2
The zeros of 30=67t-5t^2 are t=(67-sqrt(3889))/(10),t=(67+sqrt(3889))/(10)