What are the solutions to the equation 3x^2-6x+1=3 ?

The solutions to the equation 3x^2-6x+1=3 are x=(3+sqrt(15))/3 ,x=(3-sqrt(15))/3

Find the zeros of 3x^2-6x+1=3

The zeros of 3x^2-6x+1=3 are x=(3+sqrt(15))/3 ,x=(3-sqrt(15))/3

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3 x^{2} - 6 x + 1 = 3

x = \frac{3 + 15}{3}, x = \frac{3 - 15}{3}

Decimal Notation

x = 2.29099 \dots, x = - 0.29099 \dots

Solution steps

3 x^{2} - 6 x + 1 = 3

Move

3

to the left side

3 x^{2} - 6 x - 2 = 0

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 3 (- 2) = 215

x_{1, 2} = \frac{- ( - 6 ) \pm 2 15}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 15}{2 \cdot 3}, x_{2} = \frac{- ( - 6 ) - 2 15}{2 \cdot 3}

x = \frac{- ( - 6 ) + 2 15}{2 \cdot 3} : \frac{3 + 15}{3}

x = \frac{- ( - 6 ) - 2 15}{2 \cdot 3} : \frac{3 - 15}{3}

The solutions to the quadratic equation are:

x = \frac{3 + 15}{3}, x = \frac{3 - 15}{3}

λ^{2} + 3 λ + 2 = 0

145 x^{2} - 118 x + 41 = 0

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

x, x^{2} + y^{2} - 16 x + 2 y + 65 = 0

3 y^{2} - y = 0

What are the solutions to the equation 3x^2-6x+1=3 ?
The solutions to the equation 3x^2-6x+1=3 are x=(3+sqrt(15))/3 ,x=(3-sqrt(15))/3
Find the zeros of 3x^2-6x+1=3
The zeros of 3x^2-6x+1=3 are x=(3+sqrt(15))/3 ,x=(3-sqrt(15))/3