What are the solutions to the equation 68.2=39.2t-4.9t^2 ?

The solutions to the equation 68.2=39.2t-4.9t^2 are t=(28-sqrt(102))/7 ,t=(28+sqrt(102))/7

Find the zeros of 68.2=39.2t-4.9t^2

The zeros of 68.2=39.2t-4.9t^2 are t=(28-sqrt(102))/7 ,t=(28+sqrt(102))/7

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68.2 = 39.2 t - 4.9 t^{2}

t = \frac{28 - 102}{7}, t = \frac{28 + 102}{7}

Solution steps

68.2 = 39.2 t - 4.9 t^{2}

Multiply both sides by

10

682 = 392 t - 49 t^{2}

Switch sides

392 t - 49 t^{2} = 682

Move

682

to the left side

392 t - 49 t^{2} - 682 = 0

Write in the standard form

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

- 49 t^{2} + 392 t - 682 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 392 \pm 39 2 ^{2} - 4 ( - 49 ) ( - 682 )}{2 ( - 49 )}

39 2^{2} - 4 (- 49) (- 682) = 14102

t_{1, 2} = \frac{- 392 \pm 14 102}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 392 + 14 102}{2 ( - 49 )}, t_{2} = \frac{- 392 - 14 102}{2 ( - 49 )}

t = \frac{- 392 + 14 102}{2 ( - 49 )} : \frac{28 - 102}{7}

t = \frac{- 392 - 14 102}{2 ( - 49 )} : \frac{28 + 102}{7}

The solutions to the quadratic equation are:

t = \frac{28 - 102}{7}, t = \frac{28 + 102}{7}

(36 - R)^{2} + (18 - R)^{2} = R^{2}

5 x^{2} - 20 x - 12 = 0

4 x^{2} + x - \frac{1}{16} = 0

9 x^{2} - 6 x + 1 = 0

(x - 2)^{2} - 5 = 0

What are the solutions to the equation 68.2=39.2t-4.9t^2 ?
The solutions to the equation 68.2=39.2t-4.9t^2 are t=(28-sqrt(102))/7 ,t=(28+sqrt(102))/7
Find the zeros of 68.2=39.2t-4.9t^2
The zeros of 68.2=39.2t-4.9t^2 are t=(28-sqrt(102))/7 ,t=(28+sqrt(102))/7