What are the solutions to the equation 100=112+96t-16t^2 ?

The solutions to the equation 100=112+96t-16t^2 are t=-(-6+sqrt(39))/2 ,t=(6+sqrt(39))/2

Find the zeros of 100=112+96t-16t^2

The zeros of 100=112+96t-16t^2 are t=-(-6+sqrt(39))/2 ,t=(6+sqrt(39))/2

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100 = 112 + 96 t - 16 t^{2}

t = - \frac{- 6 + 39}{2}, t = \frac{6 + 39}{2}

Decimal Notation

t = - 0.12249 \dots, t = 6.12249 \dots

Solution steps

100 = 112 + 96 t - 16 t^{2}

Switch sides

112 + 96 t - 16 t^{2} = 100

Move

100

to the left side

- 16 t^{2} + 96 t + 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 ( - 16 ) \cdot 12}{2 ( - 16 )}

9 6^{2} - 4 (- 16) \cdot 12 = 1639

t_{1, 2} = \frac{- 96 \pm 16 39}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 96 + 16 39}{2 ( - 16 )}, t_{2} = \frac{- 96 - 16 39}{2 ( - 16 )}

t = \frac{- 96 + 16 39}{2 ( - 16 )} : - \frac{- 6 + 39}{2}

t = \frac{- 96 - 16 39}{2 ( - 16 )} : \frac{6 + 39}{2}

The solutions to the quadratic equation are:

t = - \frac{- 6 + 39}{2}, t = \frac{6 + 39}{2}

- 2 p = - 4

2 x^{2} + 3 x - 9 \leq 0

\frac{696000}{1.49 \cdot 1 0 ^{8}}

y - 10 < 0

What are the solutions to the equation 100=112+96t-16t^2 ?
The solutions to the equation 100=112+96t-16t^2 are t=-(-6+sqrt(39))/2 ,t=(6+sqrt(39))/2
Find the zeros of 100=112+96t-16t^2
The zeros of 100=112+96t-16t^2 are t=-(-6+sqrt(39))/2 ,t=(6+sqrt(39))/2