What are the solutions to the equation 2x^2-2x+9=0 ?

The solutions to the equation 2x^2-2x+9=0 are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2

Find the zeros of 2x^2-2x+9=0

The zeros of 2x^2-2x+9=0 are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2

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2 x^{2} - 2 x + 9 = 0

x = \frac{1}{2} + i \frac{17}{2}, x = \frac{1}{2} - i \frac{17}{2}

Solution steps

2 x^{2} - 2 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 9 : 217 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 17 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 17 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 17 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 17 i}{2 \cdot 2} : \frac{1}{2} + i \frac{17}{2}

x = \frac{- ( - 2 ) - 2 17 i}{2 \cdot 2} : \frac{1}{2} - i \frac{17}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{17}{2}, x = \frac{1}{2} - i \frac{17}{2}

- 2 y + 7 < 1 or 4 y + 3 \leq - 5

- 8 x + 6 \leq 2 (- 3 x + 1)

∣ 6 x + 42 ∣ = 0

10 x^{3} - 80

\frac{- 3}{- 1 - 5}

What are the solutions to the equation 2x^2-2x+9=0 ?
The solutions to the equation 2x^2-2x+9=0 are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2
Find the zeros of 2x^2-2x+9=0
The zeros of 2x^2-2x+9=0 are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2