What are the solutions to the equation 34=162+32t-16t^2 ?

The solutions to the equation 34=162+32t-16t^2 are t=-2,t=4

Find the zeros of 34=162+32t-16t^2

The zeros of 34=162+32t-16t^2 are t=-2,t=4

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34 = 162 + 32 t - 16 t^{2}

t = - 2, t = 4

Solution steps

34 = 162 + 32 t - 16 t^{2}

Switch sides

162 + 32 t - 16 t^{2} = 34

Move

34

to the left side

- 16 t^{2} + 32 t + 128 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) \cdot 128}{2 ( - 16 )}

3 2^{2} - 4 (- 16) \cdot 128 = 96

t_{1, 2} = \frac{- 32 \pm 96}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 32 + 96}{2 ( - 16 )}, t_{2} = \frac{- 32 - 96}{2 ( - 16 )}

t = \frac{- 32 + 96}{2 ( - 16 )} : - 2

t = \frac{- 32 - 96}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = - 2, t = 4

9 x^{2} + 9 x + 2

z^{2} - 9 z + 20

2 > - 2

x^{3} - 4 x^{2} < 0

∣ v ∣ = 0

What are the solutions to the equation 34=162+32t-16t^2 ?
The solutions to the equation 34=162+32t-16t^2 are t=-2,t=4
Find the zeros of 34=162+32t-16t^2
The zeros of 34=162+32t-16t^2 are t=-2,t=4