What are the solutions to the equation -16t^2+12t+40=0 ?

The solutions to the equation -16t^2+12t+40=0 are t=-5/4 ,t=2

Find the zeros of -16t^2+12t+40=0

The zeros of -16t^2+12t+40=0 are t=-5/4 ,t=2

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- 16 t^{2} + 12 t + 40 = 0

t = - \frac{5}{4}, t = 2

Decimal Notation

t = - 1.25, t = 2

Solution steps

- 16 t^{2} + 12 t + 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 16 ) \cdot 40}{2 ( - 16 )}

1 2^{2} - 4 (- 16) \cdot 40 = 52

t_{1, 2} = \frac{- 12 \pm 52}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 12 + 52}{2 ( - 16 )}, t_{2} = \frac{- 12 - 52}{2 ( - 16 )}

t = \frac{- 12 + 52}{2 ( - 16 )} : - \frac{5}{4}

t = \frac{- 12 - 52}{2 ( - 16 )} : 2

The solutions to the quadratic equation are:

t = - \frac{5}{4}, t = 2

- 5 (\frac{1}{2})

12 + 9

a^{2} - b^{2} + 3 a + 3 b

5 t - 3 < 12 or - 3 t + 4 \geq 16

7 t - 44 + t^{2}

What are the solutions to the equation -16t^2+12t+40=0 ?
The solutions to the equation -16t^2+12t+40=0 are t=-5/4 ,t=2
Find the zeros of -16t^2+12t+40=0
The zeros of -16t^2+12t+40=0 are t=-5/4 ,t=2