What are the solutions to the equation 14(x-4)-(x+2)=(x+2)(x-4) ?

The solutions to the equation 14(x-4)-(x+2)=(x+2)(x-4) are x=10,x=5

Find the zeros of 14(x-4)-(x+2)=(x+2)(x-4)

The zeros of 14(x-4)-(x+2)=(x+2)(x-4) are x=10,x=5

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14 (x - 4) - (x + 2) = (x + 2) (x - 4)

x = 10, x = 5

Solution steps

14 (x - 4) - (x + 2) = (x + 2) (x - 4)

Expand

14 (x - 4) - (x + 2) : 13 x - 58

Expand

(x + 2) (x - 4) : x^{2} - 2 x - 8

13 x - 58 = x^{2} - 2 x - 8

Switch sides

x^{2} - 2 x - 8 = 13 x - 58

Move

58

to the left side

x^{2} - 2 x + 50 = 13 x

Move

13 x

to the left side

x^{2} - 15 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 50}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 50 = 5

x_{1, 2} = \frac{- ( - 15 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 15 ) - 5}{2 \cdot 1}

x = \frac{- ( - 15 ) + 5}{2 \cdot 1} : 10

x = \frac{- ( - 15 ) - 5}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

x = 10, x = 5

6 a^{2} - 36 a - 162

9 a^{3} x^{2} - 18 a x^{3}

k x + p x - k y - p y

6 x^{2} + 72 x + 120 = 0

- 2 (3) + 2

What are the solutions to the equation 14(x-4)-(x+2)=(x+2)(x-4) ?
The solutions to the equation 14(x-4)-(x+2)=(x+2)(x-4) are x=10,x=5
Find the zeros of 14(x-4)-(x+2)=(x+2)(x-4)
The zeros of 14(x-4)-(x+2)=(x+2)(x-4) are x=10,x=5