What are the solutions to the equation 8x^2-x+4=0 ?

The solutions to the equation 8x^2-x+4=0 are x= 1/16+i(sqrt(127))/(16),x= 1/16-i(sqrt(127))/(16)

Find the zeros of 8x^2-x+4=0

The zeros of 8x^2-x+4=0 are x= 1/16+i(sqrt(127))/(16),x= 1/16-i(sqrt(127))/(16)

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8 x^{2} - x + 4 = 0

x = \frac{1}{16} + i \frac{127}{16}, x = \frac{1}{16} - i \frac{127}{16}

Solution steps

8 x^{2} - x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 8 \cdot 4 : 127 i

x_{1, 2} = \frac{- ( - 1 ) \pm 127 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 127 i}{2 \cdot 8}, x_{2} = \frac{- ( - 1 ) - 127 i}{2 \cdot 8}

x = \frac{- ( - 1 ) + 127 i}{2 \cdot 8} : \frac{1}{16} + i \frac{127}{16}

x = \frac{- ( - 1 ) - 127 i}{2 \cdot 8} : \frac{1}{16} - i \frac{127}{16}

The solutions to the quadratic equation are:

x = \frac{1}{16} + i \frac{127}{16}, x = \frac{1}{16} - i \frac{127}{16}

\frac{a + 13}{10} - \frac{2 a}{5} = \frac{3 - a}{15} + \frac{a}{2}

∣ 4 (x - 1) + 10 ∣ \leq 14

(2^{8})^{\frac{1}{2}}

0 - 4

2 x^{2} + 11 x + 9 = 10 x + 10

What are the solutions to the equation 8x^2-x+4=0 ?
The solutions to the equation 8x^2-x+4=0 are x= 1/16+i(sqrt(127))/(16),x= 1/16-i(sqrt(127))/(16)
Find the zeros of 8x^2-x+4=0
The zeros of 8x^2-x+4=0 are x= 1/16+i(sqrt(127))/(16),x= 1/16-i(sqrt(127))/(16)