What are the solutions to the equation -16t^2+48t+160=160 ?

The solutions to the equation -16t^2+48t+160=160 are t=0,t=3

Find the zeros of -16t^2+48t+160=160

The zeros of -16t^2+48t+160=160 are t=0,t=3

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

t = 0, t = 3

Solution steps

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

Move

160

to the left side

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

Solve with the quadratic formula

t_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

4 8^{2} - 4 (- 16) \cdot 0 = 48

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

Separate the solutions

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

t = \frac{- 48 + 48}{2 ( - 16 )} : 0

t = \frac{- 48 - 48}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

t = 0, t = 3

r (r + 4) = 0

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

x^{2} - 13 + 40 = 0

- x^{2} + 2 x = 3

0.0057 v^{2} + 0.13 v - 100 = 0

What are the solutions to the equation -16t^2+48t+160=160 ?
The solutions to the equation -16t^2+48t+160=160 are t=0,t=3
Find the zeros of -16t^2+48t+160=160
The zeros of -16t^2+48t+160=160 are t=0,t=3