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

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

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

The zeros of (x+2)\times (x-4)=x+2 are x=5,x=-2

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(x + 2) \cdot (x - 4) = x + 2

x = 5, x = - 2

Solution steps

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

Expand

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

x^{2} - 2 x - 8 = x + 2

Move

2

to the left side

x^{2} - 2 x - 10 = x

Move

x

to the left side

x^{2} - 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 7}{2 \cdot 1}

x = \frac{- ( - 3 ) + 7}{2 \cdot 1} : 5

x = \frac{- ( - 3 ) - 7}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 5, x = - 2

2 x + 15 = - \frac{4}{5} x^{2} + 10 x

34 y^{2} - 835 y + 4875 = 0

7 x - 4 = 2 x^{2}

x^{2} - 6 x = - 9

40 = x^{2}

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