What are the solutions to the equation 8x-4x^2=-40/3 ?

The solutions to the equation 8x-4x^2=-40/3 are x=-(-3+sqrt(39))/3 ,x=(3+sqrt(39))/3

Find the zeros of 8x-4x^2=-40/3

The zeros of 8x-4x^2=-40/3 are x=-(-3+sqrt(39))/3 ,x=(3+sqrt(39))/3

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8 x - 4 x^{2} = - \frac{40}{3}

x = - \frac{- 3 + 39}{3}, x = \frac{3 + 39}{3}

Solution steps

8 x - 4 x^{2} = - \frac{40}{3}

Multiply both sides by

3

24 x - 12 x^{2} = - 40

Move

40

to the left side

24 x - 12 x^{2} + 40 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 12 x^{2} + 24 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 ( - 12 ) \cdot 40}{2 ( - 12 )}

2 4^{2} - 4 (- 12) \cdot 40 = 839

x_{1, 2} = \frac{- 24 \pm 8 39}{2 ( - 12 )}

Separate the solutions

x_{1} = \frac{- 24 + 8 39}{2 ( - 12 )}, x_{2} = \frac{- 24 - 8 39}{2 ( - 12 )}

x = \frac{- 24 + 8 39}{2 ( - 12 )} : - \frac{- 3 + 39}{3}

x = \frac{- 24 - 8 39}{2 ( - 12 )} : \frac{3 + 39}{3}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 39}{3}, x = \frac{3 + 39}{3}

x - \frac{x + 1}{2} = 3

20 y^{2} - 5 x^{2}

\frac{1}{9} x^{4} + 4 x^{2} + 36

8 x^{5} y^{3} z^{3} + 60 x^{3} y^{6} z - 4 x^{8} y^{4}

(\frac{1}{2})^{4} \cdot (\frac{5}{2})^{- 2}

What are the solutions to the equation 8x-4x^2=-40/3 ?
The solutions to the equation 8x-4x^2=-40/3 are x=-(-3+sqrt(39))/3 ,x=(3+sqrt(39))/3
Find the zeros of 8x-4x^2=-40/3
The zeros of 8x-4x^2=-40/3 are x=-(-3+sqrt(39))/3 ,x=(3+sqrt(39))/3