What are the solutions to the equation (x+1)(x+2)=3 ?

The solutions to the equation (x+1)(x+2)=3 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2

Find the zeros of (x+1)(x+2)=3

The zeros of (x+1)(x+2)=3 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 1) (x + 2) = 3

x = \frac{- 3 + 13}{2}, x = \frac{- 3 - 13}{2}

Decimal Notation

x = 0.30277 \dots, x = - 3.30277 \dots

Solution steps

(x + 1) (x + 2) = 3

Expand

(x + 1) (x + 2) : x^{2} + 3 x + 2

x^{2} + 3 x + 2 = 3

Move

3

to the left side

x^{2} + 3 x - 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

x_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 13}{2 \cdot 1}, x_{2} = \frac{- 3 - 13}{2 \cdot 1}

x = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

x = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 13}{2}, x = \frac{- 3 - 13}{2}

x^{2} + 32 = 9 x

4 m^{2} - 3 m - 6 = 0

(2 x - 1) (x + 3) = x + 1

x^{2} + 11 x + 20 = 0

2 a^{2} + 2 a - 60 = 0

What are the solutions to the equation (x+1)(x+2)=3 ?
The solutions to the equation (x+1)(x+2)=3 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2
Find the zeros of (x+1)(x+2)=3
The zeros of (x+1)(x+2)=3 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2