What are the solutions to the equation 3x^2-2x+7=0 ?

The solutions to the equation 3x^2-2x+7=0 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3

Find the zeros of 3x^2-2x+7=0

The zeros of 3x^2-2x+7=0 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3

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3 x^{2} - 2 x + 7 = 0

x = \frac{1}{3} + i \frac{2 5}{3}, x = \frac{1}{3} - i \frac{2 5}{3}

Solution steps

3 x^{2} - 2 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

(- 2)^{2} - 4 \cdot 3 \cdot 7 : 45 i

x_{1, 2} = \frac{- ( - 2 ) \pm 4 5 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 5 i}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 4 5 i}{2 \cdot 3}

x = \frac{- ( - 2 ) + 4 5 i}{2 \cdot 3} : \frac{1}{3} + i \frac{2 5}{3}

x = \frac{- ( - 2 ) - 4 5 i}{2 \cdot 3} : \frac{1}{3} - i \frac{2 5}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{2 5}{3}, x = \frac{1}{3} - i \frac{2 5}{3}

k - 7 = - 12

\frac{1}{2} + \frac{1}{4} + \frac{1}{8}

\frac{3}{4} n + 1 = - 5

9 s + 3 = 57

3 r^{2} + 5 r - 1 = 0

What are the solutions to the equation 3x^2-2x+7=0 ?
The solutions to the equation 3x^2-2x+7=0 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3
Find the zeros of 3x^2-2x+7=0
The zeros of 3x^2-2x+7=0 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3