What are the solutions to the equation 3(7x-2)+3x(x-4)=1 ?

The solutions to the equation 3(7x-2)+3x(x-4)=1 are x=(-9+sqrt(165))/6 ,x=(-9-sqrt(165))/6

Find the zeros of 3(7x-2)+3x(x-4)=1

The zeros of 3(7x-2)+3x(x-4)=1 are x=(-9+sqrt(165))/6 ,x=(-9-sqrt(165))/6

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3 (7 x - 2) + 3 x (x - 4) = 1

x = \frac{- 9 + 165}{6}, x = \frac{- 9 - 165}{6}

Decimal Notation

x = 0.64087 \dots, x = - 3.64087 \dots

Solution steps

3 (7 x - 2) + 3 x (x - 4) = 1

Expand

3 (7 x - 2) + 3 x (x - 4) : 3 x^{2} + 9 x - 6

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

Move

1

to the left side

3 x^{2} + 9 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

9^{2} - 4 \cdot 3 (- 7) = 165

x_{1, 2} = \frac{- 9 \pm 165}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 9 + 165}{2 \cdot 3}, x_{2} = \frac{- 9 - 165}{2 \cdot 3}

x = \frac{- 9 + 165}{2 \cdot 3} : \frac{- 9 + 165}{6}

x = \frac{- 9 - 165}{2 \cdot 3} : \frac{- 9 - 165}{6}

The solutions to the quadratic equation are:

x = \frac{- 9 + 165}{6}, x = \frac{- 9 - 165}{6}

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

(k + 7)^{2} = 289

x^{2} + 2 = 6 x + 4

x (x - 2) = 24

15 x^{2} - 75 x + 90 = 0

What are the solutions to the equation 3(7x-2)+3x(x-4)=1 ?
The solutions to the equation 3(7x-2)+3x(x-4)=1 are x=(-9+sqrt(165))/6 ,x=(-9-sqrt(165))/6
Find the zeros of 3(7x-2)+3x(x-4)=1
The zeros of 3(7x-2)+3x(x-4)=1 are x=(-9+sqrt(165))/6 ,x=(-9-sqrt(165))/6