What are the solutions to the equation (3x+2)(4x+1)-3=0 ?

The solutions to the equation (3x+2)(4x+1)-3=0 are x= 1/12 ,x=-1

Find the zeros of (3x+2)(4x+1)-3=0

The zeros of (3x+2)(4x+1)-3=0 are x= 1/12 ,x=-1

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(3 x + 2) (4 x + 1) - 3 = 0

x = \frac{1}{12}, x = - 1

Decimal Notation

x = 0.08333 \dots, x = - 1

Solution steps

(3 x + 2) (4 x + 1) - 3 = 0

Expand

(3 x + 2) (4 x + 1) - 3 : 12 x^{2} + 11 x - 1

12 x^{2} + 11 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 12 ( - 1 )}{2 \cdot 12}

1 1^{2} - 4 \cdot 12 (- 1) = 13

x_{1, 2} = \frac{- 11 \pm 13}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 11 + 13}{2 \cdot 12}, x_{2} = \frac{- 11 - 13}{2 \cdot 12}

x = \frac{- 11 + 13}{2 \cdot 12} : \frac{1}{12}

x = \frac{- 11 - 13}{2 \cdot 12} : - 1

The solutions to the quadratic equation are:

x = \frac{1}{12}, x = - 1

\frac{11.7}{\frac{1}{2} ( 4.6 ) ^{2}} \cdot 4.6

9 + 3 x \leq 24

\frac{4}{5} - \frac{1}{8}

6 a^{2} - a - 5 = 0

2 + 3 x > 7

What are the solutions to the equation (3x+2)(4x+1)-3=0 ?
The solutions to the equation (3x+2)(4x+1)-3=0 are x= 1/12 ,x=-1
Find the zeros of (3x+2)(4x+1)-3=0
The zeros of (3x+2)(4x+1)-3=0 are x= 1/12 ,x=-1