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

The solutions to the equation (x+3)(x-2)=1 are x=(-1+sqrt(29))/2 ,x=(-1-sqrt(29))/2

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

The zeros of (x+3)(x-2)=1 are x=(-1+sqrt(29))/2 ,x=(-1-sqrt(29))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{- 1 + 29}{2}, x = \frac{- 1 - 29}{2}

Decimal Notation

x = 2.19258 \dots, x = - 3.19258 \dots

Solution steps

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

Expand

(x + 3) (x - 2) : x^{2} + x - 6

x^{2} + x - 6 = 1

Move

1

to the left side

x^{2} + x - 7 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 7) = 29

x_{1, 2} = \frac{- 1 \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 29}{2 \cdot 1}, x_{2} = \frac{- 1 - 29}{2 \cdot 1}

x = \frac{- 1 + 29}{2 \cdot 1} : \frac{- 1 + 29}{2}

x = \frac{- 1 - 29}{2 \cdot 1} : \frac{- 1 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 29}{2}, x = \frac{- 1 - 29}{2}

x^{2} + 10 x = 3

4 x^{2} + 5 = 54

4 x^{2} + 5 = 41

4^{2} + b^{2} = 1 0^{2}

3 x - 5 x^{2} + 1 = x - 7

What are the solutions to the equation (x+3)(x-2)=1 ?
The solutions to the equation (x+3)(x-2)=1 are x=(-1+sqrt(29))/2 ,x=(-1-sqrt(29))/2
Find the zeros of (x+3)(x-2)=1
The zeros of (x+3)(x-2)=1 are x=(-1+sqrt(29))/2 ,x=(-1-sqrt(29))/2