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

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

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

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

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

t = - 1, t = 3

Solution steps

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

Write in the standard form

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

- 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

0 = \frac{1}{2} x^{2}

k^{2} + 10 k + 10 = 9

c^{2} + 5 c - 24 = 0

18 x^{2} + 24 x = - 8

12 x^{2} - 27 = 0

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