What are the solutions to the equation 0=40+5t-4.9t^2 ?

The solutions to the equation 0=40+5t-4.9t^2 are t=-(5(sqrt(809)-5))/(49),t=(5(5+sqrt(809)))/(49)

Find the zeros of 0=40+5t-4.9t^2

The zeros of 0=40+5t-4.9t^2 are t=-(5(sqrt(809)-5))/(49),t=(5(5+sqrt(809)))/(49)

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0 = 40 + 5 t - 4.9 t^{2}

t = - \frac{5 ( 809 - 5 )}{49}, t = \frac{5 ( 5 + 809 )}{49}

Solution steps

0 = 40 + 5 t - 4.9 t^{2}

Multiply both sides by

10

0 = 400 + 50 t - 49 t^{2}

Switch sides

400 + 50 t - 49 t^{2} = 0

Write in the standard form

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

- 49 t^{2} + 50 t + 400 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 ( - 49 ) \cdot 400}{2 ( - 49 )}

5 0^{2} - 4 (- 49) \cdot 400 = 10809

t_{1, 2} = \frac{- 50 \pm 10 809}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 50 + 10 809}{2 ( - 49 )}, t_{2} = \frac{- 50 - 10 809}{2 ( - 49 )}

t = \frac{- 50 + 10 809}{2 ( - 49 )} : - \frac{5 ( 809 - 5 )}{49}

t = \frac{- 50 - 10 809}{2 ( - 49 )} : \frac{5 ( 5 + 809 )}{49}

The solutions to the quadratic equation are:

t = - \frac{5 ( 809 - 5 )}{49}, t = \frac{5 ( 5 + 809 )}{49}

4 x \cdot (2 x + 4) = - 8

2 x^{2} - 7 x = 12

6 x^{2} + 8 = - 16

x^{2} - 2 x - 1 = - 1

2 q^{2} + q - 1596 = 0

What are the solutions to the equation 0=40+5t-4.9t^2 ?
The solutions to the equation 0=40+5t-4.9t^2 are t=-(5(sqrt(809)-5))/(49),t=(5(5+sqrt(809)))/(49)
Find the zeros of 0=40+5t-4.9t^2
The zeros of 0=40+5t-4.9t^2 are t=-(5(sqrt(809)-5))/(49),t=(5(5+sqrt(809)))/(49)