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

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

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

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

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

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

Decimal Notation

x = 3.63299 \dots, x = 0.36700 \dots

Solution steps

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

Switch sides

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}

2 x^{2} + 4 x = 0

x, 12 z^{2} = 4 y^{2} - x^{2}

\frac{x ^{2} - 5 x + 4}{8 - 5} = 5

3 x^{2} - 5 x = 2 x^{2} + 4 x + 10

5 x^{2} + 12 x - 44 = 0

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