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

The solutions to the equation 48=-16t^2+32t+48 are t=0,t=2

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

The zeros of 48=-16t^2+32t+48 are t=0,t=2

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

t = 0, t = 2

Solution steps

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

Switch sides

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

Move

48

to the left side

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

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

3 2^{2} - 4 (- 16) \cdot 0 = 32

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

Separate the solutions

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

t = \frac{- 32 + 32}{2 ( - 16 )} : 0

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

The solutions to the quadratic equation are:

t = 0, t = 2

\frac{8 + 192}{4}

\frac{18}{x + 6} \leq 3

a c - a d + b c - b d

9 x = 7 x + 8

(4 \cdot 3)^{\frac{2}{3}}

What are the solutions to the equation 48=-16t^2+32t+48 ?
The solutions to the equation 48=-16t^2+32t+48 are t=0,t=2
Find the zeros of 48=-16t^2+32t+48
The zeros of 48=-16t^2+32t+48 are t=0,t=2