What are the solutions to the equation 62=-4.9t^2+19.6t+58.8 ?

The solutions to the equation 62=-4.9t^2+19.6t+58.8 are t=(2(7-sqrt(41)))/7 ,t=(2(7+sqrt(41)))/7

Find the zeros of 62=-4.9t^2+19.6t+58.8

The zeros of 62=-4.9t^2+19.6t+58.8 are t=(2(7-sqrt(41)))/7 ,t=(2(7+sqrt(41)))/7

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62 = - 4.9 t^{2} + 19.6 t + 58.8

t = \frac{2 ( 7 - 41 )}{7}, t = \frac{2 ( 7 + 41 )}{7}

Decimal Notation

t = 0.17053 \dots, t = 3.82946 \dots

Solution steps

62 = - 4.9 t^{2} + 19.6 t + 58.8

Multiply both sides by

10

620 = - 49 t^{2} + 196 t + 588

Switch sides

- 49 t^{2} + 196 t + 588 = 620

Move

620

to the left side

- 49 t^{2} + 196 t - 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 196 \pm 19 6 ^{2} - 4 ( - 49 ) ( - 32 )}{2 ( - 49 )}

19 6^{2} - 4 (- 49) (- 32) = 2841

t_{1, 2} = \frac{- 196 \pm 28 41}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 196 + 28 41}{2 ( - 49 )}, t_{2} = \frac{- 196 - 28 41}{2 ( - 49 )}

t = \frac{- 196 + 28 41}{2 ( - 49 )} : \frac{2 ( 7 - 41 )}{7}

t = \frac{- 196 - 28 41}{2 ( - 49 )} : \frac{2 ( 7 + 41 )}{7}

The solutions to the quadratic equation are:

t = \frac{2 ( 7 - 41 )}{7}, t = \frac{2 ( 7 + 41 )}{7}

t^{2} - 81 = 0

x, x^{2} + 4 x + 3 = 0

2 x^{2} + 19 x + 9 = 0

(4 b + 1)^{2} = 12

2 s^{2} + 6 s + 4 = 0

What are the solutions to the equation 62=-4.9t^2+19.6t+58.8 ?
The solutions to the equation 62=-4.9t^2+19.6t+58.8 are t=(2(7-sqrt(41)))/7 ,t=(2(7+sqrt(41)))/7
Find the zeros of 62=-4.9t^2+19.6t+58.8
The zeros of 62=-4.9t^2+19.6t+58.8 are t=(2(7-sqrt(41)))/7 ,t=(2(7+sqrt(41)))/7