What are the solutions to the equation -16t^2+48t+64=0 ?

The solutions to the equation -16t^2+48t+64=0 are t=-1,t=4

Find the zeros of -16t^2+48t+64=0

The zeros of -16t^2+48t+64=0 are t=-1,t=4

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- 16 t^{2} + 48 t + 64 = 0

t = - 1, t = 4

Solution steps

- 16 t^{2} + 48 t + 64 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) \cdot 64}{2 ( - 16 )}

4 8^{2} - 4 (- 16) \cdot 64 = 80

t_{1, 2} = \frac{- 48 \pm 80}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 48 + 80}{2 ( - 16 )}, t_{2} = \frac{- 48 - 80}{2 ( - 16 )}

t = \frac{- 48 + 80}{2 ( - 16 )} : - 1

t = \frac{- 48 - 80}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = - 1, t = 4

lo g_{\frac{1}{3}} (x) > lo g_{x} (3 - \frac{5}{2})

3 x^{2} + 18 x + 24 \geq 0

x^{2} - 2 x = 6

p = 3.51 - 50 + (48) e^{- 0.05 (0.25)}

∣ x + 5 ∣ = 5

What are the solutions to the equation -16t^2+48t+64=0 ?
The solutions to the equation -16t^2+48t+64=0 are t=-1,t=4
Find the zeros of -16t^2+48t+64=0
The zeros of -16t^2+48t+64=0 are t=-1,t=4