What are the solutions to the equation (2x+1)(3x-1)=6 ?

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

Find the zeros of (2x+1)(3x-1)=6

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

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(2 x + 1) (3 x - 1) = 6

x = 1, x = - \frac{7}{6}

Decimal Notation

x = 1, x = - 1.16666 \dots

Solution steps

(2 x + 1) (3 x - 1) = 6

Expand

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

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

Move

6

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 6 (- 7) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 6}, x_{2} = \frac{- 1 - 13}{2 \cdot 6}

x = \frac{- 1 + 13}{2 \cdot 6} : 1

x = \frac{- 1 - 13}{2 \cdot 6} : - \frac{7}{6}

The solutions to the quadratic equation are:

x = 1, x = - \frac{7}{6}

1 1^{2} + x^{2} = 1 6^{2}

x^{2} + 8 x + 11 = 0

4.9 x^{2} - 30.88 x + 4 = 0

x^{2} = 2.7 x

2 (x - 5)^{2} + 13 = 31

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