What are the solutions to the equation 32-32x=(6x+4)(x-1) ?

The solutions to the equation 32-32x=(6x+4)(x-1) are x=1,x=-6

Find the zeros of 32-32x=(6x+4)(x-1)

The zeros of 32-32x=(6x+4)(x-1) are x=1,x=-6

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32 - 32 x = (6 x + 4) (x - 1)

x = 1, x = - 6

Solution steps

32 - 32 x = (6 x + 4) (x - 1)

Expand

(6 x + 4) (x - 1) : 6 x^{2} - 2 x - 4

32 - 32 x = 6 x^{2} - 2 x - 4

Switch sides

6 x^{2} - 2 x - 4 = 32 - 32 x

Move

32 x

to the left side

6 x^{2} + 30 x - 4 = 32

Move

32

to the left side

6 x^{2} + 30 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 \cdot 6 ( - 36 )}{2 \cdot 6}

3 0^{2} - 4 \cdot 6 (- 36) = 42

x_{1, 2} = \frac{- 30 \pm 42}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 30 + 42}{2 \cdot 6}, x_{2} = \frac{- 30 - 42}{2 \cdot 6}

x = \frac{- 30 + 42}{2 \cdot 6} : 1

x = \frac{- 30 - 42}{2 \cdot 6} : - 6

The solutions to the quadratic equation are:

x = 1, x = - 6

\frac{86204}{461.4 \cdot 294.2}

\frac{2 x ^{2}}{3 x - 2} \leq - 1

\frac{3}{5} = \frac{3}{10} d - 12

\frac{x - 2}{4} \geq 5

(x - 2) (1 - 5 (x - \frac{1}{5})) < - 5 (x^{2} - 2)

What are the solutions to the equation 32-32x=(6x+4)(x-1) ?
The solutions to the equation 32-32x=(6x+4)(x-1) are x=1,x=-6
Find the zeros of 32-32x=(6x+4)(x-1)
The zeros of 32-32x=(6x+4)(x-1) are x=1,x=-6