What are the solutions to the equation solvefor x,(x+5)(x+2)=(x+2) ?

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

Find the zeros of solvefor x,(x+5)(x+2)=(x+2)

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

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x, (x + 5) (x + 2) = (x + 2)

x = - 2, x = - 4

Solution steps

(x + 5) (x + 2) = (x + 2)

Expand

(x + 5) (x + 2) : x^{2} + 7 x + 10

x^{2} + 7 x + 10 = x + 2

Move

2

to the left side

x^{2} + 7 x + 8 = x

Move

x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 8 = 2

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

Separate the solutions

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

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

x = \frac{- 6 - 2}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = - 2, x = - 4

v, v + 3 \cdot 1 0^{- 3} (v - 1)^{2} 10 \cdot 1 0^{3} = 5

(3 x - 4) (3 x - 4) = 8^{8}

39 = \frac{1}{2} (h - 7) h

(52)^{2} = 1^{2} + x^{2}

x^{2} + 15 x + 21 = 0

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