What are the solutions to the equation 0=-16t^2+32t+8 ?

The solutions to the equation 0=-16t^2+32t+8 are t=-(-2+sqrt(6))/2 ,t=(2+sqrt(6))/2

Find the zeros of 0=-16t^2+32t+8

The zeros of 0=-16t^2+32t+8 are t=-(-2+sqrt(6))/2 ,t=(2+sqrt(6))/2

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0 = - 16 t^{2} + 32 t + 8

t = - \frac{- 2 + 6}{2}, t = \frac{2 + 6}{2}

Solution steps

0 = - 16 t^{2} + 32 t + 8

Switch sides

- 16 t^{2} + 32 t + 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) \cdot 8}{2 ( - 16 )}

3 2^{2} - 4 (- 16) \cdot 8 = 166

t_{1, 2} = \frac{- 32 \pm 16 6}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 32 + 16 6}{2 ( - 16 )}, t_{2} = \frac{- 32 - 16 6}{2 ( - 16 )}

t = \frac{- 32 + 16 6}{2 ( - 16 )} : - \frac{- 2 + 6}{2}

t = \frac{- 32 - 16 6}{2 ( - 16 )} : \frac{2 + 6}{2}

The solutions to the quadratic equation are:

t = - \frac{- 2 + 6}{2}, t = \frac{2 + 6}{2}

x^{2} + 4 x - 96 = 0

a^{7}

n^{2} + 3 n - 54

15 r^{2} + 7 r - 4 = 0

20 - 4

What are the solutions to the equation 0=-16t^2+32t+8 ?
The solutions to the equation 0=-16t^2+32t+8 are t=-(-2+sqrt(6))/2 ,t=(2+sqrt(6))/2
Find the zeros of 0=-16t^2+32t+8
The zeros of 0=-16t^2+32t+8 are t=-(-2+sqrt(6))/2 ,t=(2+sqrt(6))/2