What are the solutions to the equation -16t^2+47.2t+3=7 ?

The solutions to the equation -16t^2+47.2t+3=7 are t=(59-sqrt(3081))/(40),t=(59+sqrt(3081))/(40)

Find the zeros of -16t^2+47.2t+3=7

The zeros of -16t^2+47.2t+3=7 are t=(59-sqrt(3081))/(40),t=(59+sqrt(3081))/(40)

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- 16 t^{2} + 47.2 t + 3 = 7

t = \frac{59 - 3081}{40}, t = \frac{59 + 3081}{40}

Solution steps

- 16 t^{2} + 47.2 t + 3 = 7

Multiply both sides by

10

- 160 t^{2} + 472 t + 30 = 70

Move

70

to the left side

- 160 t^{2} + 472 t - 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 472 \pm 47 2 ^{2} - 4 ( - 160 ) ( - 40 )}{2 ( - 160 )}

47 2^{2} - 4 (- 160) (- 40) = 83081

t_{1, 2} = \frac{- 472 \pm 8 3081}{2 ( - 160 )}

Separate the solutions

t_{1} = \frac{- 472 + 8 3081}{2 ( - 160 )}, t_{2} = \frac{- 472 - 8 3081}{2 ( - 160 )}

t = \frac{- 472 + 8 3081}{2 ( - 160 )} : \frac{59 - 3081}{40}

t = \frac{- 472 - 8 3081}{2 ( - 160 )} : \frac{59 + 3081}{40}

The solutions to the quadratic equation are:

t = \frac{59 - 3081}{40}, t = \frac{59 + 3081}{40}

\frac{2}{3} \frac{1}{2}

\frac{4 x ^{2} - 4 x - 8}{- 4 x + 8} \leq 0

4 x^{2} - 8 x + 63 = 0

2 a^{2} - 3 a - 2

x^{2} = 27

What are the solutions to the equation -16t^2+47.2t+3=7 ?
The solutions to the equation -16t^2+47.2t+3=7 are t=(59-sqrt(3081))/(40),t=(59+sqrt(3081))/(40)
Find the zeros of -16t^2+47.2t+3=7
The zeros of -16t^2+47.2t+3=7 are t=(59-sqrt(3081))/(40),t=(59+sqrt(3081))/(40)