What are the solutions to the equation 0=2x^2-x+1 ?

The solutions to the equation 0=2x^2-x+1 are x= 1/4+i(sqrt(7))/4 ,x= 1/4-i(sqrt(7))/4

Find the zeros of 0=2x^2-x+1

The zeros of 0=2x^2-x+1 are x= 1/4+i(sqrt(7))/4 ,x= 1/4-i(sqrt(7))/4

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0 = 2 x^{2} - x + 1

x = \frac{1}{4} + i \frac{7}{4}, x = \frac{1}{4} - i \frac{7}{4}

Solution steps

0 = 2 x^{2} - x + 1

Switch sides

2 x^{2} - x + 1 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 7 i}{2 \cdot 2} : \frac{1}{4} + i \frac{7}{4}

x = \frac{- ( - 1 ) - 7 i}{2 \cdot 2} : \frac{1}{4} - i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{7}{4}, x = \frac{1}{4} - i \frac{7}{4}

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

y^{2} - 3 y = 2

- x^{2} + 3 = x^{2} + 1

16 x^{2} + 9 x - 12 = 4 x + 9 x^{2}

2 x^{2} + 2 x = 24

What are the solutions to the equation 0=2x^2-x+1 ?
The solutions to the equation 0=2x^2-x+1 are x= 1/4+i(sqrt(7))/4 ,x= 1/4-i(sqrt(7))/4
Find the zeros of 0=2x^2-x+1
The zeros of 0=2x^2-x+1 are x= 1/4+i(sqrt(7))/4 ,x= 1/4-i(sqrt(7))/4