What are the solutions to the equation 30=-4.9t^2+28t+40 ?

The solutions to the equation 30=-4.9t^2+28t+40 are t=-(10(sqrt(5)-2))/7 ,t=(10(2+sqrt(5)))/7

Find the zeros of 30=-4.9t^2+28t+40

The zeros of 30=-4.9t^2+28t+40 are t=-(10(sqrt(5)-2))/7 ,t=(10(2+sqrt(5)))/7

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30 = - 4.9 t^{2} + 28 t + 40

t = - \frac{10 ( 5 - 2 )}{7}, t = \frac{10 ( 2 + 5 )}{7}

Decimal Notation

t = - 0.33723 \dots, t = 6.05152 \dots

Solution steps

30 = - 4.9 t^{2} + 28 t + 40

Multiply both sides by

10

300 = - 49 t^{2} + 280 t + 400

Switch sides

- 49 t^{2} + 280 t + 400 = 300

Move

300

to the left side

- 49 t^{2} + 280 t + 100 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 280 \pm 28 0 ^{2} - 4 ( - 49 ) \cdot 100}{2 ( - 49 )}

28 0^{2} - 4 (- 49) \cdot 100 = 1405

t_{1, 2} = \frac{- 280 \pm 140 5}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 280 + 140 5}{2 ( - 49 )}, t_{2} = \frac{- 280 - 140 5}{2 ( - 49 )}

t = \frac{- 280 + 140 5}{2 ( - 49 )} : - \frac{10 ( 5 - 2 )}{7}

t = \frac{- 280 - 140 5}{2 ( - 49 )} : \frac{10 ( 2 + 5 )}{7}

The solutions to the quadratic equation are:

t = - \frac{10 ( 5 - 2 )}{7}, t = \frac{10 ( 2 + 5 )}{7}

- 32 (4)

\frac{6 ! 7 !}{4 ! 9 !}

3 x - x + 2 > 4 - 6

lo g_{4} (3) = n - 2

What are the solutions to the equation 30=-4.9t^2+28t+40 ?
The solutions to the equation 30=-4.9t^2+28t+40 are t=-(10(sqrt(5)-2))/7 ,t=(10(2+sqrt(5)))/7
Find the zeros of 30=-4.9t^2+28t+40
The zeros of 30=-4.9t^2+28t+40 are t=-(10(sqrt(5)-2))/7 ,t=(10(2+sqrt(5)))/7