What are the solutions to the equation -16t^2+32t+240=0 ?

The solutions to the equation -16t^2+32t+240=0 are t=-3,t=5

Find the zeros of -16t^2+32t+240=0

The zeros of -16t^2+32t+240=0 are t=-3,t=5

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- 16 t^{2} + 32 t + 240 = 0

t = - 3, t = 5

Solution steps

- 16 t^{2} + 32 t + 240 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) \cdot 240}{2 ( - 16 )}

3 2^{2} - 4 (- 16) \cdot 240 = 128

t_{1, 2} = \frac{- 32 \pm 128}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 32 + 128}{2 ( - 16 )}, t_{2} = \frac{- 32 - 128}{2 ( - 16 )}

t = \frac{- 32 + 128}{2 ( - 16 )} : - 3

t = \frac{- 32 - 128}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = - 3, t = 5

3 (2 x - 7) = 6 (x - 4) - 3

\frac{6 - ( - 7 )}{- 3 - ( - 7 )}

28 a + 14

g (- 2) = 2 (- 2) + 1

3 - x^{2} = 7 - 5 x

What are the solutions to the equation -16t^2+32t+240=0 ?
The solutions to the equation -16t^2+32t+240=0 are t=-3,t=5
Find the zeros of -16t^2+32t+240=0
The zeros of -16t^2+32t+240=0 are t=-3,t=5