You may use the quadratic formula such that:

`x_(1,2) = (-3+-sqrt(3^2 - 4*1*2))/2`

`x_(1,2) = (-3+-sqrt(9-8))/2`

`x_(1,2) = (-3+-1)/2`

`x_1 = (-3+1)/2 =gt x_1 = -1`

`x_2 = (-3-1)/2 =gt x_2 = -2`

**Hence, evaluating solutions to equation `x^2 + 3x + 2 = 0 ` yields `x_1 = -1` and `x_2 = -2` .**

yamero,

I suppose that you missed to put that equation is equal to zero.

To solve the question above, you should first factor the equation to (X+2)(X+1)=0

then X can be either -2 or -1 so that the value in the parenthesis becomes zero.

x^2 + 3x + 2

You can solve by considering it as a quadratic equation x^2 + 3x + 2=0

i.e. **ax^2+bx+c=0**

**Where a=1**

** b=3**

** c=2**

Quadratic Formula,

= [-3+sqrt{3^2-4(1)(2)}]/2(1) =[-3-sqrt{3^2-4(1)(2)}]/2(1)

= [-3+sqrt{9-8}]/2 =[-3-sqrt{9-8}]/2

= [-3+sqrt{1}]/2 =[-3-sqrt{1}]/2

= [-3+1]/2 =[-3-1]/2

= -2/2 =-4/2

** x= -1 x=-2**

x^2 + 3x + 2

(x+2) (x+1)

x = -2 or -1

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