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

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

Find the zeros of (3x+2)(x-1)=3x

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

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(3 x + 2) (x - 1) = 3 x

x = \frac{2 + 10}{3}, x = \frac{2 - 10}{3}

Decimal Notation

x = 1.72075 \dots, x = - 0.38742 \dots

Solution steps

(3 x + 2) (x - 1) = 3 x

Expand

(3 x + 2) (x - 1) : 3 x^{2} - x - 2

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

Move

3 x

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 4 ) \pm 2 10}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 10}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 10}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 10}{2 \cdot 3} : \frac{2 + 10}{3}

x = \frac{- ( - 4 ) - 2 10}{2 \cdot 3} : \frac{2 - 10}{3}

The solutions to the quadratic equation are:

x = \frac{2 + 10}{3}, x = \frac{2 - 10}{3}

6 x^{2} + x + 1 = 0

2 k^{2} - 37 = 61

x^{2} - 7 = 9

y^{2} + 6 y

- 3 x^{2} + 297 = 0

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