What are the solutions to the equation 3x^2-12x+1=0 ?

The solutions to the equation 3x^2-12x+1=0 are x=(6+sqrt(33))/3 ,x=(6-sqrt(33))/3

Find the zeros of 3x^2-12x+1=0

The zeros of 3x^2-12x+1=0 are x=(6+sqrt(33))/3 ,x=(6-sqrt(33))/3

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3 x^{2} - 12 x + 1 = 0

x = \frac{6 + 33}{3}, x = \frac{6 - 33}{3}

Decimal Notation

x = 3.91485 \dots, x = 0.08514 \dots

Solution steps

3 x^{2} - 12 x + 1 = 0

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 3 \cdot 1 = 233

x_{1, 2} = \frac{- ( - 12 ) \pm 2 33}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 33}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 2 33}{2 \cdot 3}

x = \frac{- ( - 12 ) + 2 33}{2 \cdot 3} : \frac{6 + 33}{3}

x = \frac{- ( - 12 ) - 2 33}{2 \cdot 3} : \frac{6 - 33}{3}

The solutions to the quadratic equation are:

x = \frac{6 + 33}{3}, x = \frac{6 - 33}{3}

360 \div 2 \div 3 \div 5 \div 2

n^{2} - 18 n + 56

3 a^{4} + 16 a^{2} + 20

x^{2} + \frac{19 x}{2} + \frac{9}{2} = 0

\frac{( x - 4 ) ( x + 5 )}{x - 6} \leq 0

What are the solutions to the equation 3x^2-12x+1=0 ?
The solutions to the equation 3x^2-12x+1=0 are x=(6+sqrt(33))/3 ,x=(6-sqrt(33))/3
Find the zeros of 3x^2-12x+1=0
The zeros of 3x^2-12x+1=0 are x=(6+sqrt(33))/3 ,x=(6-sqrt(33))/3