What are the solutions to the equation 50=-16t^2+48t+210 ?

The solutions to the equation 50=-16t^2+48t+210 are t=-2,t=5

Find the zeros of 50=-16t^2+48t+210

The zeros of 50=-16t^2+48t+210 are t=-2,t=5

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50 = - 16 t^{2} + 48 t + 210

t = - 2, t = 5

Solution steps

50 = - 16 t^{2} + 48 t + 210

Switch sides

- 16 t^{2} + 48 t + 210 = 50

Move

50

to the left side

- 16 t^{2} + 48 t + 160 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) \cdot 160}{2 ( - 16 )}

4 8^{2} - 4 (- 16) \cdot 160 = 112

t_{1, 2} = \frac{- 48 \pm 112}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 48 + 112}{2 ( - 16 )}, t_{2} = \frac{- 48 - 112}{2 ( - 16 )}

t = \frac{- 48 + 112}{2 ( - 16 )} : - 2

t = \frac{- 48 - 112}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = - 2, t = 5

64 b^{2} - 25

2 x^{4} - 7 x^{3} + 12 x^{2} - 8 x - 8 = 0

6 (- y + 4) - 3 (2 y - 3) = - y + 5 + 5 y

3 m^{2} + 6 m + 3

x^{3} (x^{2})

What are the solutions to the equation 50=-16t^2+48t+210 ?
The solutions to the equation 50=-16t^2+48t+210 are t=-2,t=5
Find the zeros of 50=-16t^2+48t+210
The zeros of 50=-16t^2+48t+210 are t=-2,t=5