What are the solutions to the equation 96=80t-16t^2 ?

The solutions to the equation 96=80t-16t^2 are t=2,t=3

Find the zeros of 96=80t-16t^2

The zeros of 96=80t-16t^2 are t=2,t=3

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96 = 80 t - 16 t^{2}

t = 2, t = 3

Solution steps

96 = 80 t - 16 t^{2}

Switch sides

80 t - 16 t^{2} = 96

Move

96

to the left side

80 t - 16 t^{2} - 96 = 0

Write in the standard form

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

- 16 t^{2} + 80 t - 96 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) ( - 96 )}{2 ( - 16 )}

8 0^{2} - 4 (- 16) (- 96) = 16

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

Separate the solutions

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

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

t = \frac{- 80 - 16}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

t = 2, t = 3

x^{2} + 4 x + 4 = 0

8^{2} = 5^{2} + x^{2}

2 1^{2} + b^{2} = 2 9^{2}

- 2 x^{2} - 7 x + 15 = 0

x^{2} + 4 x + 17 = 8 - 2 x

What are the solutions to the equation 96=80t-16t^2 ?
The solutions to the equation 96=80t-16t^2 are t=2,t=3
Find the zeros of 96=80t-16t^2
The zeros of 96=80t-16t^2 are t=2,t=3