What are the solutions to the equation 12=35t-5t^2 ?

The solutions to the equation 12=35t-5t^2 are t=(35-sqrt(985))/(10),t=(35+sqrt(985))/(10)

Find the zeros of 12=35t-5t^2

The zeros of 12=35t-5t^2 are t=(35-sqrt(985))/(10),t=(35+sqrt(985))/(10)

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12 = 35 t - 5 t^{2}

t = \frac{35 - 985}{10}, t = \frac{35 + 985}{10}

Decimal Notation

t = 0.36152 \dots, t = 6.63847 \dots

Solution steps

12 = 35 t - 5 t^{2}

Switch sides

35 t - 5 t^{2} = 12

Move

12

to the left side

35 t - 5 t^{2} - 12 = 0

Write in the standard form

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

- 5 t^{2} + 35 t - 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 35 \pm 3 5 ^{2} - 4 ( - 5 ) ( - 12 )}{2 ( - 5 )}

3 5^{2} - 4 (- 5) (- 12) = 985

t_{1, 2} = \frac{- 35 \pm 985}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 35 + 985}{2 ( - 5 )}, t_{2} = \frac{- 35 - 985}{2 ( - 5 )}

t = \frac{- 35 + 985}{2 ( - 5 )} : \frac{35 - 985}{10}

t = \frac{- 35 - 985}{2 ( - 5 )} : \frac{35 + 985}{10}

The solutions to the quadratic equation are:

t = \frac{35 - 985}{10}, t = \frac{35 + 985}{10}

1 0^{- 3.3}

- 10 x + 25 \leq 5

12 x^{2} - 56 x + 9

x^{\frac{1}{2}} \cdot x^{\frac{2}{3}}

\frac{6 + i}{- 3 - 2 i}

What are the solutions to the equation 12=35t-5t^2 ?
The solutions to the equation 12=35t-5t^2 are t=(35-sqrt(985))/(10),t=(35+sqrt(985))/(10)
Find the zeros of 12=35t-5t^2
The zeros of 12=35t-5t^2 are t=(35-sqrt(985))/(10),t=(35+sqrt(985))/(10)