What are the solutions to the equation -16t^2+34t+42=0 ?

The solutions to the equation -16t^2+34t+42=0 are t=-7/8 ,t=3

Find the zeros of -16t^2+34t+42=0

The zeros of -16t^2+34t+42=0 are t=-7/8 ,t=3

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- 16 t^{2} + 34 t + 42 = 0

t = - \frac{7}{8}, t = 3

Decimal Notation

t = - 0.875, t = 3

Solution steps

- 16 t^{2} + 34 t + 42 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 34 \pm 3 4 ^{2} - 4 ( - 16 ) \cdot 42}{2 ( - 16 )}

3 4^{2} - 4 (- 16) \cdot 42 = 62

t_{1, 2} = \frac{- 34 \pm 62}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 34 + 62}{2 ( - 16 )}, t_{2} = \frac{- 34 - 62}{2 ( - 16 )}

t = \frac{- 34 + 62}{2 ( - 16 )} : - \frac{7}{8}

t = \frac{- 34 - 62}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

t = - \frac{7}{8}, t = 3

x^{2} - 8 (x + 3) = + 4 x - 45

2 x^{2} + 3 x - 54 = 0

- 16 x^{2} - 8 x + 64 = 0

x^{2} + 22 x + 116.5 = 0

4 x^{2} - 36 x = - 69

What are the solutions to the equation -16t^2+34t+42=0 ?
The solutions to the equation -16t^2+34t+42=0 are t=-7/8 ,t=3
Find the zeros of -16t^2+34t+42=0
The zeros of -16t^2+34t+42=0 are t=-7/8 ,t=3