What are the solutions to the equation 12=2+27t-16t^2 ?

The solutions to the equation 12=2+27t-16t^2 are t=(27-sqrt(89))/(32),t=(27+sqrt(89))/(32)

Find the zeros of 12=2+27t-16t^2

The zeros of 12=2+27t-16t^2 are t=(27-sqrt(89))/(32),t=(27+sqrt(89))/(32)

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12 = 2 + 27 t - 16 t^{2}

t = \frac{27 - 89}{32}, t = \frac{27 + 89}{32}

Solution steps

12 = 2 + 27 t - 16 t^{2}

Switch sides

2 + 27 t - 16 t^{2} = 12

Move

12

to the left side

- 16 t^{2} + 27 t - 10 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 ( - 16 ) ( - 10 )}{2 ( - 16 )}

2 7^{2} - 4 (- 16) (- 10) = 89

t_{1, 2} = \frac{- 27 \pm 89}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 27 + 89}{2 ( - 16 )}, t_{2} = \frac{- 27 - 89}{2 ( - 16 )}

t = \frac{- 27 + 89}{2 ( - 16 )} : \frac{27 - 89}{32}

t = \frac{- 27 - 89}{2 ( - 16 )} : \frac{27 + 89}{32}

The solutions to the quadratic equation are:

t = \frac{27 - 89}{32}, t = \frac{27 + 89}{32}

5^{2} + b^{2} = 6^{2}

15 x \cdot 25 (8000 - x) = 161000

2 x^{2} - x = 10

(x - 8) (x - 6) = 0

- 12 x \leq 60

What are the solutions to the equation 12=2+27t-16t^2 ?
The solutions to the equation 12=2+27t-16t^2 are t=(27-sqrt(89))/(32),t=(27+sqrt(89))/(32)
Find the zeros of 12=2+27t-16t^2
The zeros of 12=2+27t-16t^2 are t=(27-sqrt(89))/(32),t=(27+sqrt(89))/(32)