What are the solutions to the equation (3x-1)^2=9x+3 ?

The solutions to the equation (3x-1)^2=9x+3 are x=(5+sqrt(33))/6 ,x=(5-sqrt(33))/6

Find the zeros of (3x-1)^2=9x+3

The zeros of (3x-1)^2=9x+3 are x=(5+sqrt(33))/6 ,x=(5-sqrt(33))/6

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(3 x - 1)^{2} = 9 x + 3

x = \frac{5 + 33}{6}, x = \frac{5 - 33}{6}

Decimal

x = 1.79076 \dots, x = - 0.12409 \dots

Solution steps

(3 x - 1)^{2} = 9 x + 3

Expand

(3 x - 1)^{2} : 9 x^{2} - 6 x + 1

9 x^{2} - 6 x + 1 = 9 x + 3

Move

3

to the left side

9 x^{2} - 6 x - 2 = 9 x

Move

9 x

to the left side

9 x^{2} - 15 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

(- 15)^{2} - 4 \cdot 9 (- 2) = 333

x_{1, 2} = \frac{- ( - 15 ) \pm 3 33}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 3 33}{2 \cdot 9}, x_{2} = \frac{- ( - 15 ) - 3 33}{2 \cdot 9}

x = \frac{- ( - 15 ) + 3 33}{2 \cdot 9} : \frac{5 + 33}{6}

x = \frac{- ( - 15 ) - 3 33}{2 \cdot 9} : \frac{5 - 33}{6}

The solutions to the quadratic equation are:

x = \frac{5 + 33}{6}, x = \frac{5 - 33}{6}

x^{2} + 20 x = - 75

4 z^{2} - 4 z - 15 = 0

x^{2} + 7 x = x

2 x^{2} + 4 = 8 x

s^{2} + 9

What are the solutions to the equation (3x-1)^2=9x+3 ?
The solutions to the equation (3x-1)^2=9x+3 are x=(5+sqrt(33))/6 ,x=(5-sqrt(33))/6
Find the zeros of (3x-1)^2=9x+3
The zeros of (3x-1)^2=9x+3 are x=(5+sqrt(33))/6 ,x=(5-sqrt(33))/6