What are the solutions to the equation 5x^2-4x+1=0 ?

The solutions to the equation 5x^2-4x+1=0 are x= 2/5+i 1/5 ,x= 2/5-i 1/5

Find the zeros of 5x^2-4x+1=0

The zeros of 5x^2-4x+1=0 are x= 2/5+i 1/5 ,x= 2/5-i 1/5

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5 x^{2} - 4 x + 1 = 0

x = \frac{2}{5} + i \frac{1}{5}, x = \frac{2}{5} - i \frac{1}{5}

Solution steps

5 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 5 \cdot 1 : 2 i

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 i}{2 \cdot 5} : \frac{2}{5} + i \frac{1}{5}

x = \frac{- ( - 4 ) - 2 i}{2 \cdot 5} : \frac{2}{5} - i \frac{1}{5}

The solutions to the quadratic equation are:

x = \frac{2}{5} + i \frac{1}{5}, x = \frac{2}{5} - i \frac{1}{5}

\frac{3}{10} \cdot \frac{2}{9}

(x - 2)^{2} \leq x^{2} + 1

4 + (- 2)

lo g_{b} (\frac{x ^{3} y}{z ^{7}})

x^{2} - x + 81 = 0

What are the solutions to the equation 5x^2-4x+1=0 ?
The solutions to the equation 5x^2-4x+1=0 are x= 2/5+i 1/5 ,x= 2/5-i 1/5
Find the zeros of 5x^2-4x+1=0
The zeros of 5x^2-4x+1=0 are x= 2/5+i 1/5 ,x= 2/5-i 1/5