What are the solutions to the equation -16t^2+32t+48=0 ?

The solutions to the equation -16t^2+32t+48=0 are t=-1,t=3

Find the zeros of -16t^2+32t+48=0

The zeros of -16t^2+32t+48=0 are t=-1,t=3

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- 16 t^{2} + 32 t + 48 = 0

t = - 1, t = 3

Solution steps

- 16 t^{2} + 32 t + 48 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) \cdot 48}{2 ( - 16 )}

3 2^{2} - 4 (- 16) \cdot 48 = 64

t_{1, 2} = \frac{- 32 \pm 64}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 32 + 64}{2 ( - 16 )}, t_{2} = \frac{- 32 - 64}{2 ( - 16 )}

t = \frac{- 32 + 64}{2 ( - 16 )} : - 1

t = \frac{- 32 - 64}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

t = - 1, t = 3

2 \cdot \frac{5}{2}

\frac{2}{3 - x} \geq \frac{5}{7 - x}

\frac{- 8}{2 ( 1 )}

2 x^{2} - 7 x = 4

x^{2} \leq 100

What are the solutions to the equation -16t^2+32t+48=0 ?
The solutions to the equation -16t^2+32t+48=0 are t=-1,t=3
Find the zeros of -16t^2+32t+48=0
The zeros of -16t^2+32t+48=0 are t=-1,t=3