What are the solutions to the equation 3x^2-2x+1=4 ?

The solutions to the equation 3x^2-2x+1=4 are x=(1+sqrt(10))/3 ,x=(1-sqrt(10))/3

Find the zeros of 3x^2-2x+1=4

The zeros of 3x^2-2x+1=4 are x=(1+sqrt(10))/3 ,x=(1-sqrt(10))/3

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3 x^{2} - 2 x + 1 = 4

x = \frac{1 + 10}{3}, x = \frac{1 - 10}{3}

Solution steps

3 x^{2} - 2 x + 1 = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 3 (- 3) = 210

x_{1, 2} = \frac{- ( - 2 ) \pm 2 10}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 10}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 10}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 10}{2 \cdot 3} : \frac{1 + 10}{3}

x = \frac{- ( - 2 ) - 2 10}{2 \cdot 3} : \frac{1 - 10}{3}

The solutions to the quadratic equation are:

x = \frac{1 + 10}{3}, x = \frac{1 - 10}{3}

x^{2} - 18 x + 16 = 0

x, x^{2} - 8 x + 16 = 0

(x - 4)^{2} + (x - 1) \cdot (x + 1) = 25

x^{2} + 3 = - 24

- x^{2} + 5 x + 7 = 0

What are the solutions to the equation 3x^2-2x+1=4 ?
The solutions to the equation 3x^2-2x+1=4 are x=(1+sqrt(10))/3 ,x=(1-sqrt(10))/3
Find the zeros of 3x^2-2x+1=4
The zeros of 3x^2-2x+1=4 are x=(1+sqrt(10))/3 ,x=(1-sqrt(10))/3