What are the solutions to the equation -16t^2+5t+5=0 ?

The solutions to the equation -16t^2+5t+5=0 are t=-(-5+sqrt(345))/(32),t=(5+sqrt(345))/(32)

Find the zeros of -16t^2+5t+5=0

The zeros of -16t^2+5t+5=0 are t=-(-5+sqrt(345))/(32),t=(5+sqrt(345))/(32)

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- 16 t^{2} + 5 t + 5 = 0

t = - \frac{- 5 + 345}{32}, t = \frac{5 + 345}{32}

Decimal Notation

t = - 0.42419 \dots, t = 0.73669 \dots

Solution steps

- 16 t^{2} + 5 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 16 ) \cdot 5}{2 ( - 16 )}

5^{2} - 4 (- 16) \cdot 5 = 345

t_{1, 2} = \frac{- 5 \pm 345}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 5 + 345}{2 ( - 16 )}, t_{2} = \frac{- 5 - 345}{2 ( - 16 )}

t = \frac{- 5 + 345}{2 ( - 16 )} : - \frac{- 5 + 345}{32}

t = \frac{- 5 - 345}{2 ( - 16 )} : \frac{5 + 345}{32}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 345}{32}, t = \frac{5 + 345}{32}

- \frac{1}{3} h \geq 8

16 \frac{- 1}{2}

21 m^{5} n - 7 m^{4} n^{2} + 7 m^{3} n^{3} - 7 m^{2} n

2 - 7 y \leq 16 y - 13

∣ 4 x + 2 ∣ + 3 > 21

What are the solutions to the equation -16t^2+5t+5=0 ?
The solutions to the equation -16t^2+5t+5=0 are t=-(-5+sqrt(345))/(32),t=(5+sqrt(345))/(32)
Find the zeros of -16t^2+5t+5=0
The zeros of -16t^2+5t+5=0 are t=-(-5+sqrt(345))/(32),t=(5+sqrt(345))/(32)