What are the solutions to the equation 3x^2-4x+6=2 ?

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

Find the zeros of 3x^2-4x+6=2

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

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3 x^{2} - 4 x + 6 = 2

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

Solution steps

3 x^{2} - 4 x + 6 = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 4 : 42 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- \frac{2}{3} (0) - 1

[(3 - \frac{5}{4}) \cdot (- \frac{7}{6})] \div \frac{15}{2}

3 x + 19 \leq 7 x - 13

x^{3} - x^{2} - x + 1

x^{2} + 9 \leq 0

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