What are the solutions to the equation 4x(9-3x)=8x-1 ?

The solutions to the equation 4x(9-3x)=8x-1 are x=-(-7+2sqrt(13))/6 ,x=(7+2sqrt(13))/6

Find the zeros of 4x(9-3x)=8x-1

The zeros of 4x(9-3x)=8x-1 are x=-(-7+2sqrt(13))/6 ,x=(7+2sqrt(13))/6

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4 x (9 - 3 x) = 8 x - 1

x = - \frac{- 7 + 2 13}{6}, x = \frac{7 + 2 13}{6}

Solution steps

4 x (9 - 3 x) = 8 x - 1

Expand

4 x (9 - 3 x) : 36 x - 12 x^{2}

36 x - 12 x^{2} = 8 x - 1

Move

1

to the left side

36 x - 12 x^{2} + 1 = 8 x

Move

8 x

to the left side

- 12 x^{2} + 28 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 ( - 12 ) \cdot 1}{2 ( - 12 )}

2 8^{2} - 4 (- 12) \cdot 1 = 813

x_{1, 2} = \frac{- 28 \pm 8 13}{2 ( - 12 )}

Separate the solutions

x_{1} = \frac{- 28 + 8 13}{2 ( - 12 )}, x_{2} = \frac{- 28 - 8 13}{2 ( - 12 )}

x = \frac{- 28 + 8 13}{2 ( - 12 )} : - \frac{- 7 + 2 13}{6}

x = \frac{- 28 - 8 13}{2 ( - 12 )} : \frac{7 + 2 13}{6}

The solutions to the quadratic equation are:

x = - \frac{- 7 + 2 13}{6}, x = \frac{7 + 2 13}{6}

2 s^{2} + s + 3 = 0

3 - x^{2} \leq - 2 x + 4

x^{2} - \frac{5}{6} x + \frac{1}{6} = 0

1058

m^{2} + 2 m = 8

What are the solutions to the equation 4x(9-3x)=8x-1 ?
The solutions to the equation 4x(9-3x)=8x-1 are x=-(-7+2sqrt(13))/6 ,x=(7+2sqrt(13))/6
Find the zeros of 4x(9-3x)=8x-1
The zeros of 4x(9-3x)=8x-1 are x=-(-7+2sqrt(13))/6 ,x=(7+2sqrt(13))/6