What are the solutions to the equation -8x^2+2x=5 ?

The solutions to the equation -8x^2+2x=5 are x= 1/8-i(sqrt(39))/8 ,x= 1/8+i(sqrt(39))/8

Find the zeros of -8x^2+2x=5

The zeros of -8x^2+2x=5 are x= 1/8-i(sqrt(39))/8 ,x= 1/8+i(sqrt(39))/8

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- 8 x^{2} + 2 x = 5

x = \frac{1}{8} - i \frac{39}{8}, x = \frac{1}{8} + i \frac{39}{8}

Solution steps

- 8 x^{2} + 2 x = 5

Move

5

to the left side

- 8 x^{2} + 2 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 8 ) ( - 5 )}{2 ( - 8 )}

Simplify

2^{2} - 4 (- 8) (- 5) : 239 i

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

Separate the solutions

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

x = \frac{- 2 + 2 39 i}{2 ( - 8 )} : \frac{1}{8} - i \frac{39}{8}

x = \frac{- 2 - 2 39 i}{2 ( - 8 )} : \frac{1}{8} + i \frac{39}{8}

The solutions to the quadratic equation are:

x = \frac{1}{8} - i \frac{39}{8}, x = \frac{1}{8} + i \frac{39}{8}

y, z = x^{3} + 4 x y - 3 y^{2}

x, x^{2} - 3 x y + 5 y^{2} - 4 = 0

- 16 x^{2} + 144 x + 108 = 0

y, x^{3} + x y^{2} = 5

5 x^{2} - 7 x + 7 = 0

What are the solutions to the equation -8x^2+2x=5 ?
The solutions to the equation -8x^2+2x=5 are x= 1/8-i(sqrt(39))/8 ,x= 1/8+i(sqrt(39))/8
Find the zeros of -8x^2+2x=5
The zeros of -8x^2+2x=5 are x= 1/8-i(sqrt(39))/8 ,x= 1/8+i(sqrt(39))/8