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

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

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

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

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

t = - 1, t = 4

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

- 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

\frac{( 2 x ^{- 3} y ) ^{4}}{8 x ^{2} y ^{2}}

\frac{7}{2} - \frac{9}{4}

2 x \leq - 7

(0.2)^{4}

\frac{5}{6} - \frac{2}{9}

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