What are the solutions to the equation 112=112+96t-16t^2 ?

The solutions to the equation 112=112+96t-16t^2 are t=0,t=6

Find the zeros of 112=112+96t-16t^2

The zeros of 112=112+96t-16t^2 are t=0,t=6

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112 = 112 + 96 t - 16 t^{2}

t = 0, t = 6

Solution steps

112 = 112 + 96 t - 16 t^{2}

Switch sides

112 + 96 t - 16 t^{2} = 112

Move

112

to the left side

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

Solve with the quadratic formula

t_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

9 6^{2} - 4 (- 16) \cdot 0 = 96

t_{1, 2} = \frac{- 96 \pm 96}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 96 + 96}{2 ( - 16 )}, t_{2} = \frac{- 96 - 96}{2 ( - 16 )}

t = \frac{- 96 + 96}{2 ( - 16 )} : 0

t = \frac{- 96 - 96}{2 ( - 16 )} : 6

The solutions to the quadratic equation are:

t = 0, t = 6

12 x^{3} - 9 x^{2}

- 5 n - 6 n \leq 8 - 8 n - n

(x - 4)^{2} = 27

lo g_{2} (\frac{1}{16})

4 x^{2} + 20 x = 56

What are the solutions to the equation 112=112+96t-16t^2 ?
The solutions to the equation 112=112+96t-16t^2 are t=0,t=6
Find the zeros of 112=112+96t-16t^2
The zeros of 112=112+96t-16t^2 are t=0,t=6