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

The solutions to the equation (3x-2)^2=9x+2 are x=(7+sqrt(41))/6 ,x=(7-sqrt(41))/6

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

The zeros of (3x-2)^2=9x+2 are x=(7+sqrt(41))/6 ,x=(7-sqrt(41))/6

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(3 x - 2)^{2} = 9 x + 2

x = \frac{7 + 41}{6}, x = \frac{7 - 41}{6}

Decimal Notation

x = 2.23385 \dots, x = 0.09947 \dots

Solution steps

(3 x - 2)^{2} = 9 x + 2

Expand

(3 x - 2)^{2} : 9 x^{2} - 12 x + 4

9 x^{2} - 12 x + 4 = 9 x + 2

Move

2

to the left side

9 x^{2} - 12 x + 2 = 9 x

Move

9 x

to the left side

9 x^{2} - 21 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 9 \cdot 2}{2 \cdot 9}

(- 21)^{2} - 4 \cdot 9 \cdot 2 = 341

x_{1, 2} = \frac{- ( - 21 ) \pm 3 41}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 3 41}{2 \cdot 9}, x_{2} = \frac{- ( - 21 ) - 3 41}{2 \cdot 9}

x = \frac{- ( - 21 ) + 3 41}{2 \cdot 9} : \frac{7 + 41}{6}

x = \frac{- ( - 21 ) - 3 41}{2 \cdot 9} : \frac{7 - 41}{6}

The solutions to the quadratic equation are:

x = \frac{7 + 41}{6}, x = \frac{7 - 41}{6}

x + 3 x^{2} = 0

m^{2} - 9 m + 9 = 0

7 u^{2} + 2 u - 15 = 0

3 x^{2} - 6 x + 17

x, y^{2} x - x^{2} y = y - 2 x

What are the solutions to the equation (3x-2)^2=9x+2 ?
The solutions to the equation (3x-2)^2=9x+2 are x=(7+sqrt(41))/6 ,x=(7-sqrt(41))/6
Find the zeros of (3x-2)^2=9x+2
The zeros of (3x-2)^2=9x+2 are x=(7+sqrt(41))/6 ,x=(7-sqrt(41))/6