What are the solutions to the equation 6(x-3)-6(x-4)=(x-4)(x-3) ?

The solutions to the equation 6(x-3)-6(x-4)=(x-4)(x-3) are x=6,x=1

Find the zeros of 6(x-3)-6(x-4)=(x-4)(x-3)

The zeros of 6(x-3)-6(x-4)=(x-4)(x-3) are x=6,x=1

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6 (x - 3) - 6 (x - 4) = (x - 4) (x - 3)

x = 6, x = 1

Solution steps

6 (x - 3) - 6 (x - 4) = (x - 4) (x - 3)

Expand

6 (x - 3) - 6 (x - 4) : 6

Expand

(x - 4) (x - 3) : x^{2} - 7 x + 12

6 = x^{2} - 7 x + 12

Switch sides

x^{2} - 7 x + 12 = 6

Move

6

to the left side

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

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 1 \cdot 6 = 5

x_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 1}

x = \frac{- ( - 7 ) + 5}{2 \cdot 1} : 6

x = \frac{- ( - 7 ) - 5}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 6, x = 1

80 \cdot 10 = \frac{v ^{2}}{2}

16.1 t^{2} - 200 t - 200 = 0

3 x^{2} - 6 x - 244 = 0

x^{2} - 3 x = 0

6 x (x - 1) = 35 - 7 x

What are the solutions to the equation 6(x-3)-6(x-4)=(x-4)(x-3) ?
The solutions to the equation 6(x-3)-6(x-4)=(x-4)(x-3) are x=6,x=1
Find the zeros of 6(x-3)-6(x-4)=(x-4)(x-3)
The zeros of 6(x-3)-6(x-4)=(x-4)(x-3) are x=6,x=1