What are the solutions to the equation (x+5)(x+1)=x-1 ?

The solutions to the equation (x+5)(x+1)=x-1 are x=-2,x=-3

Find the zeros of (x+5)(x+1)=x-1

The zeros of (x+5)(x+1)=x-1 are x=-2,x=-3

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(x + 5) (x + 1) = x - 1

x = - 2, x = - 3

Solution steps

(x + 5) (x + 1) = x - 1

Expand

(x + 5) (x + 1) : x^{2} + 6 x + 5

x^{2} + 6 x + 5 = x - 1

Move

1

to the left side

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

Move

x

to the left side

x^{2} + 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

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

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

Separate the solutions

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

x = \frac{- 5 + 1}{2 \cdot 1} : - 2

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

The solutions to the quadratic equation are:

x = - 2, x = - 3

x, f = I n (x^{2} + 7)

0 = x^{2} - x + 2

x, z^{3} + 3 z + y = x^{2} + 1

x^{2} - 4 x + 9

x^{2} - 4 x - 3

What are the solutions to the equation (x+5)(x+1)=x-1 ?
The solutions to the equation (x+5)(x+1)=x-1 are x=-2,x=-3
Find the zeros of (x+5)(x+1)=x-1
The zeros of (x+5)(x+1)=x-1 are x=-2,x=-3