What are the solutions to the equation -16t^2+152t+8=0 ?

The solutions to the equation -16t^2+152t+8=0 are t=-(-19+3sqrt(41))/4 ,t=(19+3sqrt(41))/4

Find the zeros of -16t^2+152t+8=0

The zeros of -16t^2+152t+8=0 are t=-(-19+3sqrt(41))/4 ,t=(19+3sqrt(41))/4

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- 16 t^{2} + 152 t + 8 = 0

t = - \frac{- 19 + 3 41}{4}, t = \frac{19 + 3 41}{4}

Decimal Notation

t = - 0.05234 \dots, t = 9.55234 \dots

Solution steps

- 16 t^{2} + 152 t + 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 152 \pm 15 2 ^{2} - 4 ( - 16 ) \cdot 8}{2 ( - 16 )}

15 2^{2} - 4 (- 16) \cdot 8 = 2441

t_{1, 2} = \frac{- 152 \pm 24 41}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 152 + 24 41}{2 ( - 16 )}, t_{2} = \frac{- 152 - 24 41}{2 ( - 16 )}

t = \frac{- 152 + 24 41}{2 ( - 16 )} : - \frac{- 19 + 3 41}{4}

t = \frac{- 152 - 24 41}{2 ( - 16 )} : \frac{19 + 3 41}{4}

The solutions to the quadratic equation are:

t = - \frac{- 19 + 3 41}{4}, t = \frac{19 + 3 41}{4}

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What are the solutions to the equation -16t^2+152t+8=0 ?
The solutions to the equation -16t^2+152t+8=0 are t=-(-19+3sqrt(41))/4 ,t=(19+3sqrt(41))/4
Find the zeros of -16t^2+152t+8=0
The zeros of -16t^2+152t+8=0 are t=-(-19+3sqrt(41))/4 ,t=(19+3sqrt(41))/4