What are the solutions to the equation 8-8x=(3x+1)(x-1) ?

The solutions to the equation 8-8x=(3x+1)(x-1) are x=1,x=-3

Find the zeros of 8-8x=(3x+1)(x-1)

The zeros of 8-8x=(3x+1)(x-1) are x=1,x=-3

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8 - 8 x = (3 x + 1) (x - 1)

x = 1, x = - 3

Solution steps

8 - 8 x = (3 x + 1) (x - 1)

Expand

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

8 - 8 x = 3 x^{2} - 2 x - 1

Switch sides

3 x^{2} - 2 x - 1 = 8 - 8 x

Move

8 x

to the left side

3 x^{2} + 6 x - 1 = 8

Move

8

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 9 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 9) = 12

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

Separate the solutions

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

x = \frac{- 6 + 12}{2 \cdot 3} : 1

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

The solutions to the quadratic equation are:

x = 1, x = - 3

\frac{( \frac{2}{x + y} )}{( \frac{x + y}{2} )}

2 (29 + 4 (5))^{- \frac{1}{2}}

4 x + 2 [4 - 2 (x + 2)] = 2 x - 4

2 x^{3} - 18 x - x^{2} y + 9 y

\frac{1}{- \frac{12}{13}}

What are the solutions to the equation 8-8x=(3x+1)(x-1) ?
The solutions to the equation 8-8x=(3x+1)(x-1) are x=1,x=-3
Find the zeros of 8-8x=(3x+1)(x-1)
The zeros of 8-8x=(3x+1)(x-1) are x=1,x=-3