What are the solutions to the equation 8x^2-x+2=0 ?

The solutions to the equation 8x^2-x+2=0 are x= 1/16+i(3sqrt(7))/(16),x= 1/16-i(3sqrt(7))/(16)

Find the zeros of 8x^2-x+2=0

The zeros of 8x^2-x+2=0 are x= 1/16+i(3sqrt(7))/(16),x= 1/16-i(3sqrt(7))/(16)

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

x = \frac{1}{16} + i \frac{3 7}{16}, x = \frac{1}{16} - i \frac{3 7}{16}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 8 \cdot 2 : 37 i

x_{1, 2} = \frac{- ( - 1 ) \pm 3 7 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 8}, x_{2} = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 8}

x = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 8} : \frac{1}{16} + i \frac{3 7}{16}

x = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 8} : \frac{1}{16} - i \frac{3 7}{16}

The solutions to the quadratic equation are:

x = \frac{1}{16} + i \frac{3 7}{16}, x = \frac{1}{16} - i \frac{3 7}{16}

7 x^{2} + 7 x + 7

(x^{- 1} + y^{- 1})^{- 2}

(5)^{- 1}

\frac{3}{4} (x + 3) = 9

∣ - 9 x ∣ = 64

What are the solutions to the equation 8x^2-x+2=0 ?
The solutions to the equation 8x^2-x+2=0 are x= 1/16+i(3sqrt(7))/(16),x= 1/16-i(3sqrt(7))/(16)
Find the zeros of 8x^2-x+2=0
The zeros of 8x^2-x+2=0 are x= 1/16+i(3sqrt(7))/(16),x= 1/16-i(3sqrt(7))/(16)