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

The solutions to the equation (x+2)(x+5)=-2 are x=-3,x=-4

Find the zeros of (x+2)(x+5)=-2

The zeros of (x+2)(x+5)=-2 are x=-3,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = - 3, x = - 4

Solution steps

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

Expand

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

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

Move

2

to the left side

x^{2} + 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

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

Separate the solutions

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

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

x = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = - 3, x = - 4

2 x^{2} - 3 x - 7 = 0

(z + 7)^{2} = - 81

\frac{4 ( 0.999 )}{( 0.999 ) - 1}

2 x - 3 > 5 or x - 3 < - 1

\frac{2 - 3}{- 2 - 5}

What are the solutions to the equation (x+2)(x+5)=-2 ?
The solutions to the equation (x+2)(x+5)=-2 are x=-3,x=-4
Find the zeros of (x+2)(x+5)=-2
The zeros of (x+2)(x+5)=-2 are x=-3,x=-4