What are the solutions to the equation 4=-16t^2+32t+6 ?

The solutions to the equation 4=-16t^2+32t+6 are t=-(-4+3sqrt(2))/4 ,t=(4+3sqrt(2))/4

Find the zeros of 4=-16t^2+32t+6

The zeros of 4=-16t^2+32t+6 are t=-(-4+3sqrt(2))/4 ,t=(4+3sqrt(2))/4

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4 = - 16 t^{2} + 32 t + 6

t = - \frac{- 4 + 3 2}{4}, t = \frac{4 + 3 2}{4}

Decimal Notation

t = - 0.06066 \dots, t = 2.06066 \dots

Solution steps

4 = - 16 t^{2} + 32 t + 6

Switch sides

- 16 t^{2} + 32 t + 6 = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

3 2^{2} - 4 (- 16) \cdot 2 = 242

t_{1, 2} = \frac{- 32 \pm 24 2}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 32 + 24 2}{2 ( - 16 )}, t_{2} = \frac{- 32 - 24 2}{2 ( - 16 )}

t = \frac{- 32 + 24 2}{2 ( - 16 )} : - \frac{- 4 + 3 2}{4}

t = \frac{- 32 - 24 2}{2 ( - 16 )} : \frac{4 + 3 2}{4}

The solutions to the quadratic equation are:

t = - \frac{- 4 + 3 2}{4}, t = \frac{4 + 3 2}{4}

x, x^{2} + y^{2} - z^{2} = 4

x, (y - x)^{2} + (z - x)^{2} = (y + x)^{2}

x^{2} - 4 x + \frac{7}{4} = 0

x^{2} - 70 = 3 x

0 = 3 x^{2} - 14 x + 15

What are the solutions to the equation 4=-16t^2+32t+6 ?
The solutions to the equation 4=-16t^2+32t+6 are t=-(-4+3sqrt(2))/4 ,t=(4+3sqrt(2))/4
Find the zeros of 4=-16t^2+32t+6
The zeros of 4=-16t^2+32t+6 are t=-(-4+3sqrt(2))/4 ,t=(4+3sqrt(2))/4