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

The solutions to the equation 6(x-1)-6(x-2)=(x-2)(x-1) are x=4,x=-1

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

The zeros of 6(x-1)-6(x-2)=(x-2)(x-1) are x=4,x=-1

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

x = 4, x = - 1

Solution steps

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

Expand

6 (x - 1) - 6 (x - 2) : 6

Expand

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

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

Switch sides

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

Move

6

to the left side

x^{2} - 3 x - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

x = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 4, x = - 1

- 15 x^{2} + 15 = 0

2 x^{2} + 0.2 x - 0.12 = 0

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

24 = 4 + 37 x - 16 x^{2}

3 x^{2} + x = 10

What are the solutions to the equation 6(x-1)-6(x-2)=(x-2)(x-1) ?
The solutions to the equation 6(x-1)-6(x-2)=(x-2)(x-1) are x=4,x=-1
Find the zeros of 6(x-1)-6(x-2)=(x-2)(x-1)
The zeros of 6(x-1)-6(x-2)=(x-2)(x-1) are x=4,x=-1