What are the solutions to the equation 3x^2-12x+4=0 ?

The solutions to the equation 3x^2-12x+4=0 are x=(2(3+sqrt(6)))/3 ,x=(2(3-sqrt(6)))/3

Find the zeros of 3x^2-12x+4=0

The zeros of 3x^2-12x+4=0 are x=(2(3+sqrt(6)))/3 ,x=(2(3-sqrt(6)))/3

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3 x^{2} - 12 x + 4 = 0

x = \frac{2 ( 3 + 6 )}{3}, x = \frac{2 ( 3 - 6 )}{3}

Decimal Notation

x = 3.63299 \dots, x = 0.36700 \dots

Solution steps

3 x^{2} - 12 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 4 = 46

x_{1, 2} = \frac{- ( - 12 ) \pm 4 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 6}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 4 6}{2 \cdot 3}

x = \frac{- ( - 12 ) + 4 6}{2 \cdot 3} : \frac{2 ( 3 + 6 )}{3}

x = \frac{- ( - 12 ) - 4 6}{2 \cdot 3} : \frac{2 ( 3 - 6 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 3 + 6 )}{3}, x = \frac{2 ( 3 - 6 )}{3}

(\frac{29}{41})^{5}

\frac{3}{2 + 1}

- 16 > + (- 10)

\frac{- 250}{- 10}

5 p^{2} - 18 p - 8

What are the solutions to the equation 3x^2-12x+4=0 ?
The solutions to the equation 3x^2-12x+4=0 are x=(2(3+sqrt(6)))/3 ,x=(2(3-sqrt(6)))/3
Find the zeros of 3x^2-12x+4=0
The zeros of 3x^2-12x+4=0 are x=(2(3+sqrt(6)))/3 ,x=(2(3-sqrt(6)))/3