What are the solutions to the equation 5x^2-6x+4=0 ?

The solutions to the equation 5x^2-6x+4=0 are x= 3/5+i(sqrt(11))/5 ,x= 3/5-i(sqrt(11))/5

Find the zeros of 5x^2-6x+4=0

The zeros of 5x^2-6x+4=0 are x= 3/5+i(sqrt(11))/5 ,x= 3/5-i(sqrt(11))/5

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5 x^{2} - 6 x + 4 = 0

x = \frac{3}{5} + i \frac{11}{5}, x = \frac{3}{5} - i \frac{11}{5}

Solution steps

5 x^{2} - 6 x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 6)^{2} - 4 \cdot 5 \cdot 4 : 211 i

x_{1, 2} = \frac{- ( - 6 ) \pm 2 11 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 11 i}{2 \cdot 5}, x_{2} = \frac{- ( - 6 ) - 2 11 i}{2 \cdot 5}

x = \frac{- ( - 6 ) + 2 11 i}{2 \cdot 5} : \frac{3}{5} + i \frac{11}{5}

x = \frac{- ( - 6 ) - 2 11 i}{2 \cdot 5} : \frac{3}{5} - i \frac{11}{5}

The solutions to the quadratic equation are:

x = \frac{3}{5} + i \frac{11}{5}, x = \frac{3}{5} - i \frac{11}{5}

- 3.4 m \leq 17

\infty^{- \infty}

\frac{1}{512} x^{9} - 512

2 x - 6

- 12 x - 8 \leq 64

What are the solutions to the equation 5x^2-6x+4=0 ?
The solutions to the equation 5x^2-6x+4=0 are x= 3/5+i(sqrt(11))/5 ,x= 3/5-i(sqrt(11))/5
Find the zeros of 5x^2-6x+4=0
The zeros of 5x^2-6x+4=0 are x= 3/5+i(sqrt(11))/5 ,x= 3/5-i(sqrt(11))/5