What are the solutions to the equation 68=140t-16t^2 ?

The solutions to the equation 68=140t-16t^2 are t=(35-sqrt(953))/8 ,t=(35+sqrt(953))/8

Find the zeros of 68=140t-16t^2

The zeros of 68=140t-16t^2 are t=(35-sqrt(953))/8 ,t=(35+sqrt(953))/8

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68 = 140 t - 16 t^{2}

t = \frac{35 - 953}{8}, t = \frac{35 + 953}{8}

Decimal Notation

t = 0.51616 \dots, t = 8.23383 \dots

Solution steps

68 = 140 t - 16 t^{2}

Switch sides

140 t - 16 t^{2} = 68

Move

68

to the left side

140 t - 16 t^{2} - 68 = 0

Write in the standard form

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

- 16 t^{2} + 140 t - 68 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 140 \pm 14 0 ^{2} - 4 ( - 16 ) ( - 68 )}{2 ( - 16 )}

14 0^{2} - 4 (- 16) (- 68) = 4953

t_{1, 2} = \frac{- 140 \pm 4 953}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 140 + 4 953}{2 ( - 16 )}, t_{2} = \frac{- 140 - 4 953}{2 ( - 16 )}

t = \frac{- 140 + 4 953}{2 ( - 16 )} : \frac{35 - 953}{8}

t = \frac{- 140 - 4 953}{2 ( - 16 )} : \frac{35 + 953}{8}

The solutions to the quadratic equation are:

t = \frac{35 - 953}{8}, t = \frac{35 + 953}{8}

448

9^{2} + 9^{2} = c^{2}

y^{2} + 13 = 6 y

4 = 0

5 x + 20 < - 3 (x - 5)

What are the solutions to the equation 68=140t-16t^2 ?
The solutions to the equation 68=140t-16t^2 are t=(35-sqrt(953))/8 ,t=(35+sqrt(953))/8
Find the zeros of 68=140t-16t^2
The zeros of 68=140t-16t^2 are t=(35-sqrt(953))/8 ,t=(35+sqrt(953))/8