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

The solutions to the equation 32=16+32t-16t^2 are t=1

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

The zeros of 32=16+32t-16t^2 are t=1

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

t = 1

Solution steps

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

Switch sides

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

Move

32

to the left side

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

Solve with the quadratic formula

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

3 2^{2} - 4 (- 16) (- 16) = 0

t_{1, 2} = \frac{- 32 \pm 0}{2 ( - 16 )}

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

\frac{- 32}{2 ( - 16 )} = 1

t = 1

The solution to the quadratic equation is:

t = 1

u 7.9 u - 2 - 3.2 u - 10 = 2.8 + u

5 x - 3 = 2 x^{2}

x^{2} + 12 x = 0

x, - 6 x y - x^{2} + 3 x = 0

- 2 x^{2} + 20 x - 42 = 0

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