What are the solutions to the equation 32=-16t^2+48t ?

The solutions to the equation 32=-16t^2+48t are t=1,t=2

Find the zeros of 32=-16t^2+48t

The zeros of 32=-16t^2+48t are t=1,t=2

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32 = - 16 t^{2} + 48 t

t = 1, t = 2

Solution steps

32 = - 16 t^{2} + 48 t

Switch sides

- 16 t^{2} + 48 t = 32

Move

32

to the left side

- 16 t^{2} + 48 t - 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) ( - 32 )}{2 ( - 16 )}

4 8^{2} - 4 (- 16) (- 32) = 16

t_{1, 2} = \frac{- 48 \pm 16}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 48 + 16}{2 ( - 16 )}, t_{2} = \frac{- 48 - 16}{2 ( - 16 )}

t = \frac{- 48 + 16}{2 ( - 16 )} : 1

t = \frac{- 48 - 16}{2 ( - 16 )} : 2

The solutions to the quadratic equation are:

t = 1, t = 2

24 x^{2} + 36 x - 24 = 0

3 x^{2} + 5 x + 2 = 0

144 = a^{2}

x, x^{2} + 2 x + 10 = 0

9 x^{2} + 12 x - 12 = 0

What are the solutions to the equation 32=-16t^2+48t ?
The solutions to the equation 32=-16t^2+48t are t=1,t=2
Find the zeros of 32=-16t^2+48t
The zeros of 32=-16t^2+48t are t=1,t=2