What are the solutions to the equation -16t^2+40t+5=0 ?

The solutions to the equation -16t^2+40t+5=0 are t=-(-5+sqrt(30))/4 ,t=(5+sqrt(30))/4

Find the zeros of -16t^2+40t+5=0

The zeros of -16t^2+40t+5=0 are t=-(-5+sqrt(30))/4 ,t=(5+sqrt(30))/4

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- 16 t^{2} + 40 t + 5 = 0

t = - \frac{- 5 + 30}{4}, t = \frac{5 + 30}{4}

Decimal Notation

t = - 0.11930 \dots, t = 2.61930 \dots

Solution steps

- 16 t^{2} + 40 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) \cdot 5}{2 ( - 16 )}

4 0^{2} - 4 (- 16) \cdot 5 = 830

t_{1, 2} = \frac{- 40 \pm 8 30}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 8 30}{2 ( - 16 )}, t_{2} = \frac{- 40 - 8 30}{2 ( - 16 )}

t = \frac{- 40 + 8 30}{2 ( - 16 )} : - \frac{- 5 + 30}{4}

t = \frac{- 40 - 8 30}{2 ( - 16 )} : \frac{5 + 30}{4}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 30}{4}, t = \frac{5 + 30}{4}

\frac{6 - 2}{- 3 + 2}

\frac{1}{x} + h

17 \leq \frac{x}{2} + 11

8 + 63

\frac{- 4 k ^{4} + 20 k ^{3}}{8 k ^{4} + 4 k ^{3}}

What are the solutions to the equation -16t^2+40t+5=0 ?
The solutions to the equation -16t^2+40t+5=0 are t=-(-5+sqrt(30))/4 ,t=(5+sqrt(30))/4
Find the zeros of -16t^2+40t+5=0
The zeros of -16t^2+40t+5=0 are t=-(-5+sqrt(30))/4 ,t=(5+sqrt(30))/4