What are the solutions to the equation 3x^2-3x+7=9x+3 ?

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

Find the zeros of 3x^2-3x+7=9x+3

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

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3 x^{2} - 3 x + 7 = 9 x + 3

x = \frac{2 ( 3 + 6 )}{3}, x = \frac{2 ( 3 - 6 )}{3}

Decimal Notation

x = 3.63299 \dots, x = 0.36700 \dots

Solution steps

3 x^{2} - 3 x + 7 = 9 x + 3

Move

3

to the left side

3 x^{2} - 3 x + 4 = 9 x

Move

9 x

to the left side

3 x^{2} - 12 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 4 = 46

x_{1, 2} = \frac{- ( - 12 ) \pm 4 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 6}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 4 6}{2 \cdot 3}

x = \frac{- ( - 12 ) + 4 6}{2 \cdot 3} : \frac{2 ( 3 + 6 )}{3}

x = \frac{- ( - 12 ) - 4 6}{2 \cdot 3} : \frac{2 ( 3 - 6 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 3 + 6 )}{3}, x = \frac{2 ( 3 - 6 )}{3}

x^{2} + 11 x + 11 = 0

u^{2} + u - 12 = 0

w^{2} - 5 w - 36 = 0

x, (x - 4) (2 x - 3) = 2 x^{2} + p x + 12

(4 ka + 4 a + 1) (k + 1) = 4 k^{2} - 7 k - 11

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