What are the solutions to the equation 7x^2-x+8=0 ?

The solutions to the equation 7x^2-x+8=0 are x= 1/14+i(sqrt(223))/(14),x= 1/14-i(sqrt(223))/(14)

Find the zeros of 7x^2-x+8=0

The zeros of 7x^2-x+8=0 are x= 1/14+i(sqrt(223))/(14),x= 1/14-i(sqrt(223))/(14)

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

x = \frac{1}{14} + i \frac{223}{14}, x = \frac{1}{14} - i \frac{223}{14}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 7 \cdot 8}{2 \cdot 7}

Simplify

(- 1)^{2} - 4 \cdot 7 \cdot 8 : 223 i

x_{1, 2} = \frac{- ( - 1 ) \pm 223 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 223 i}{2 \cdot 7}, x_{2} = \frac{- ( - 1 ) - 223 i}{2 \cdot 7}

x = \frac{- ( - 1 ) + 223 i}{2 \cdot 7} : \frac{1}{14} + i \frac{223}{14}

x = \frac{- ( - 1 ) - 223 i}{2 \cdot 7} : \frac{1}{14} - i \frac{223}{14}

The solutions to the quadratic equation are:

x = \frac{1}{14} + i \frac{223}{14}, x = \frac{1}{14} - i \frac{223}{14}

(x + 1)^{2} = 8

5 x^{2} = 27 x + 18

x, x^{2} - y^{2} - 5 x y - 1 = 0

q^{2} + 3 q - 19 = - 5

x, w = - 3 x^{2} + 5

What are the solutions to the equation 7x^2-x+8=0 ?
The solutions to the equation 7x^2-x+8=0 are x= 1/14+i(sqrt(223))/(14),x= 1/14-i(sqrt(223))/(14)
Find the zeros of 7x^2-x+8=0
The zeros of 7x^2-x+8=0 are x= 1/14+i(sqrt(223))/(14),x= 1/14-i(sqrt(223))/(14)