What are the solutions to the equation 0=-16t^2+28t+17 ?

The solutions to the equation 0=-16t^2+28t+17 are t=-(-7+3sqrt(13))/8 ,t=(7+3sqrt(13))/8

Find the zeros of 0=-16t^2+28t+17

The zeros of 0=-16t^2+28t+17 are t=-(-7+3sqrt(13))/8 ,t=(7+3sqrt(13))/8

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0 = - 16 t^{2} + 28 t + 17

t = - \frac{- 7 + 3 13}{8}, t = \frac{7 + 3 13}{8}

Decimal Notation

t = - 0.47708 \dots, t = 2.22708 \dots

Solution steps

0 = - 16 t^{2} + 28 t + 17

Switch sides

- 16 t^{2} + 28 t + 17 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 ( - 16 ) \cdot 17}{2 ( - 16 )}

2 8^{2} - 4 (- 16) \cdot 17 = 1213

t_{1, 2} = \frac{- 28 \pm 12 13}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 28 + 12 13}{2 ( - 16 )}, t_{2} = \frac{- 28 - 12 13}{2 ( - 16 )}

t = \frac{- 28 + 12 13}{2 ( - 16 )} : - \frac{- 7 + 3 13}{8}

t = \frac{- 28 - 12 13}{2 ( - 16 )} : \frac{7 + 3 13}{8}

The solutions to the quadratic equation are:

t = - \frac{- 7 + 3 13}{8}, t = \frac{7 + 3 13}{8}

(2 x - 3)^{2} = x (3 x - 1)

π r^{2} h = 200

n^{2} = - 9 - 6 n

t^{2} + 16 t - 4 = 0

\frac{n ( n + 1 )}{2} = 575

What are the solutions to the equation 0=-16t^2+28t+17 ?
The solutions to the equation 0=-16t^2+28t+17 are t=-(-7+3sqrt(13))/8 ,t=(7+3sqrt(13))/8
Find the zeros of 0=-16t^2+28t+17
The zeros of 0=-16t^2+28t+17 are t=-(-7+3sqrt(13))/8 ,t=(7+3sqrt(13))/8