What are the solutions to the equation 3x^2-4x+4=0 ?

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

Find the zeros of 3x^2-4x+4=0

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

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

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 + 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}

0 = x^{2} + 4 x + 8

∣ 2 s + 1 ∣ + 3 \leq 2

3 x (5 x + 2) = 0

2 x^{5} + 2 x^{2} 2 y^{3} - 8 x^{3} - 8 y^{3}

5^{4} \cdot 5^{- 2}

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