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

The solutions to the equation 0=5x^2-2x+6 are x= 1/5+i(sqrt(29))/5 ,x= 1/5-i(sqrt(29))/5

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

The zeros of 0=5x^2-2x+6 are x= 1/5+i(sqrt(29))/5 ,x= 1/5-i(sqrt(29))/5

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

x = \frac{1}{5} + i \frac{29}{5}, x = \frac{1}{5} - i \frac{29}{5}

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 5 \cdot 6 : 229 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 29 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 29 i}{2 \cdot 5}, x_{2} = \frac{- ( - 2 ) - 2 29 i}{2 \cdot 5}

x = \frac{- ( - 2 ) + 2 29 i}{2 \cdot 5} : \frac{1}{5} + i \frac{29}{5}

x = \frac{- ( - 2 ) - 2 29 i}{2 \cdot 5} : \frac{1}{5} - i \frac{29}{5}

The solutions to the quadratic equation are:

x = \frac{1}{5} + i \frac{29}{5}, x = \frac{1}{5} - i \frac{29}{5}

\frac{1}{2} (2)^{0}

12 - \frac{1}{2} x = x

5 x + 7 > - 13

x^{3} + 2 x^{2} - 7 x - 14

x + 4 \leq 9

What are the solutions to the equation 0=5x^2-2x+6 ?
The solutions to the equation 0=5x^2-2x+6 are x= 1/5+i(sqrt(29))/5 ,x= 1/5-i(sqrt(29))/5
Find the zeros of 0=5x^2-2x+6
The zeros of 0=5x^2-2x+6 are x= 1/5+i(sqrt(29))/5 ,x= 1/5-i(sqrt(29))/5