What are the solutions to the equation 5x^2-4x+2=0 ?

The solutions to the equation 5x^2-4x+2=0 are x= 2/5+i(sqrt(6))/5 ,x= 2/5-i(sqrt(6))/5

Find the zeros of 5x^2-4x+2=0

The zeros of 5x^2-4x+2=0 are x= 2/5+i(sqrt(6))/5 ,x= 2/5-i(sqrt(6))/5

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

x = \frac{2}{5} + i \frac{6}{5}, x = \frac{2}{5} - i \frac{6}{5}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

Simplify

(- 4)^{2} - 4 \cdot 5 \cdot 2 : 26 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 6 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 6 i}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 2 6 i}{2 \cdot 5}

x = \frac{- ( - 4 ) + 2 6 i}{2 \cdot 5} : \frac{2}{5} + i \frac{6}{5}

x = \frac{- ( - 4 ) - 2 6 i}{2 \cdot 5} : \frac{2}{5} - i \frac{6}{5}

The solutions to the quadratic equation are:

x = \frac{2}{5} + i \frac{6}{5}, x = \frac{2}{5} - i \frac{6}{5}

1 - x^{2} = 0

8 (1 + 8 x) + 8 (x - 11) < - 10 x + 2 x

(2 r^{4})^{- 5}

5 x - 3 (5 + 4 x) > 83

n^{2} - 121

What are the solutions to the equation 5x^2-4x+2=0 ?
The solutions to the equation 5x^2-4x+2=0 are x= 2/5+i(sqrt(6))/5 ,x= 2/5-i(sqrt(6))/5
Find the zeros of 5x^2-4x+2=0
The zeros of 5x^2-4x+2=0 are x= 2/5+i(sqrt(6))/5 ,x= 2/5-i(sqrt(6))/5