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

The solutions to the equation (x-1)(2x-1)=3x are x=(3+sqrt(7))/2 ,x=(3-sqrt(7))/2

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

The zeros of (x-1)(2x-1)=3x are x=(3+sqrt(7))/2 ,x=(3-sqrt(7))/2

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

x = \frac{3 + 7}{2}, x = \frac{3 - 7}{2}

Decimal Notation

x = 2.82287 \dots, x = 0.17712 \dots

Solution steps

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

Expand

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

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

Move

3 x

to the left side

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

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 2 \cdot 1 = 27

x_{1, 2} = \frac{- ( - 6 ) \pm 2 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 7}{2 \cdot 2}, x_{2} = \frac{- ( - 6 ) - 2 7}{2 \cdot 2}

x = \frac{- ( - 6 ) + 2 7}{2 \cdot 2} : \frac{3 + 7}{2}

x = \frac{- ( - 6 ) - 2 7}{2 \cdot 2} : \frac{3 - 7}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 7}{2}, x = \frac{3 - 7}{2}

6 \cdot 0

125 x^{3} + 8 y^{3}

∣ x - 1 ∣ < 3

10 < ∣ x + 5 ∣

2 (x + 4)^{2} = 90

What are the solutions to the equation (x-1)(2x-1)=3x ?
The solutions to the equation (x-1)(2x-1)=3x are x=(3+sqrt(7))/2 ,x=(3-sqrt(7))/2
Find the zeros of (x-1)(2x-1)=3x
The zeros of (x-1)(2x-1)=3x are x=(3+sqrt(7))/2 ,x=(3-sqrt(7))/2