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

The solutions to the equation 3x^2-2x+3=1+3x are x=1,x= 2/3

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

The zeros of 3x^2-2x+3=1+3x are x=1,x= 2/3

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

x = 1, x = \frac{2}{3}

Solution steps

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

Move

3 x

to the left side

3 x^{2} - 5 x + 3 = 1

Move

1

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 \cdot 2 = 1

x_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 3}

x = \frac{- ( - 5 ) + 1}{2 \cdot 3} : 1

x = \frac{- ( - 5 ) - 1}{2 \cdot 3} : \frac{2}{3}

The solutions to the quadratic equation are:

x = 1, x = \frac{2}{3}

- 2 x^{2} + 5 x + 13 = 0

2 x^{2} + 22 x + 60 = 0

(\frac{5}{6})^{6}

x^{2} = 29

e^{- 0.2496}

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