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

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

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

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

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

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

Solution steps

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

Expand

(x - 1) (2 x - 3) : 2 x^{2} - 5 x + 3

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

Move

x

to the left side

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

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 2 \cdot 3 = 23

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

10 x - 1 = 14 x - 37

cr + c t + ry + t y

2 \cdot \frac{1}{10}

3 \leq \frac{- 2 x + 3}{2} < 4

12 x^{2} + 7 x + 1 = 0

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