What are the solutions to the equation 2=x(3x-9) ?

The solutions to the equation 2=x(3x-9) are x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6

Find the zeros of 2=x(3x-9)

The zeros of 2=x(3x-9) are x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6

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2 = x (3 x - 9)

x = \frac{9 + 105}{6}, x = \frac{9 - 105}{6}

Decimal Notation

x = 3.20782 \dots, x = - 0.20782 \dots

Solution steps

2 = x (3 x - 9)

Expand

x (3 x - 9) : 3 x^{2} - 9 x

2 = 3 x^{2} - 9 x

Switch sides

3 x^{2} - 9 x = 2

Move

2

to the left side

3 x^{2} - 9 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 (- 2) = 105

x_{1, 2} = \frac{- ( - 9 ) \pm 105}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 105}{2 \cdot 3}, x_{2} = \frac{- ( - 9 ) - 105}{2 \cdot 3}

x = \frac{- ( - 9 ) + 105}{2 \cdot 3} : \frac{9 + 105}{6}

x = \frac{- ( - 9 ) - 105}{2 \cdot 3} : \frac{9 - 105}{6}

The solutions to the quadratic equation are:

x = \frac{9 + 105}{6}, x = \frac{9 - 105}{6}

- 6 x^{2} + 9 x + 8 = 0

\frac{x + x + 1}{2} = (x + 1)^{2} - 300

3 x^{2} = 12 x - 1

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

3 2^{2} = 1 6^{2} + x^{2}

What are the solutions to the equation 2=x(3x-9) ?
The solutions to the equation 2=x(3x-9) are x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6
Find the zeros of 2=x(3x-9)
The zeros of 2=x(3x-9) are x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6