What are the solutions to the equation x(3x-4)=1 ?

The solutions to the equation x(3x-4)=1 are x=(2+sqrt(7))/3 ,x=(2-sqrt(7))/3

Find the zeros of x(3x-4)=1

The zeros of x(3x-4)=1 are x=(2+sqrt(7))/3 ,x=(2-sqrt(7))/3

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x (3 x - 4) = 1

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

Decimal Notation

x = 1.54858 \dots, x = - 0.21525 \dots

Solution steps

x (3 x - 4) = 1

Expand

x (3 x - 4) : 3 x^{2} - 4 x

3 x^{2} - 4 x = 1

Move

1

to the left side

3 x^{2} - 4 x - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 3 (- 1) = 27

x_{1, 2} = \frac{- ( - 4 ) \pm 2 7}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 7}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 7}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 7}{2 \cdot 3} : \frac{2 + 7}{3}

x = \frac{- ( - 4 ) - 2 7}{2 \cdot 3} : \frac{2 - 7}{3}

The solutions to the quadratic equation are:

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

x^{2} + 0.000018 x - 0.0000018 = 0

4 x^{2} + 40 x + 280 = 0

- 9 x + 2 x^{2} - 18 = - 3 x

w^{2} - 34 w + 264 = 0

9 x^{2} + 10 x - 4 = 0

What are the solutions to the equation x(3x-4)=1 ?
The solutions to the equation x(3x-4)=1 are x=(2+sqrt(7))/3 ,x=(2-sqrt(7))/3
Find the zeros of x(3x-4)=1
The zeros of x(3x-4)=1 are x=(2+sqrt(7))/3 ,x=(2-sqrt(7))/3