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

The solutions to the equation 2x(3x+1)=7 are x=(-1+sqrt(43))/6 ,x=-(1+sqrt(43))/6

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

The zeros of 2x(3x+1)=7 are x=(-1+sqrt(43))/6 ,x=-(1+sqrt(43))/6

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2 x (3 x + 1) = 7

x = \frac{- 1 + 43}{6}, x = - \frac{1 + 43}{6}

Decimal Notation

x = 0.92623 \dots, x = - 1.25957 \dots

Solution steps

2 x (3 x + 1) = 7

Expand

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

6 x^{2} + 2 x = 7

Move

7

to the left side

6 x^{2} + 2 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 6 ( - 7 )}{2 \cdot 6}

2^{2} - 4 \cdot 6 (- 7) = 243

x_{1, 2} = \frac{- 2 \pm 2 43}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 2 + 2 43}{2 \cdot 6}, x_{2} = \frac{- 2 - 2 43}{2 \cdot 6}

x = \frac{- 2 + 2 43}{2 \cdot 6} : \frac{- 1 + 43}{6}

x = \frac{- 2 - 2 43}{2 \cdot 6} : - \frac{1 + 43}{6}

The solutions to the quadratic equation are:

x = \frac{- 1 + 43}{6}, x = - \frac{1 + 43}{6}

- \frac{2}{5} + \frac{1}{2}

5 x - 11 = 19

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

x + 6 \geq 9

4 (3 x - 2) > 12 (x + 1)

What are the solutions to the equation 2x(3x+1)=7 ?
The solutions to the equation 2x(3x+1)=7 are x=(-1+sqrt(43))/6 ,x=-(1+sqrt(43))/6
Find the zeros of 2x(3x+1)=7
The zeros of 2x(3x+1)=7 are x=(-1+sqrt(43))/6 ,x=-(1+sqrt(43))/6