quadratic formula n² +3n+2=0?

n^2 + 3n + 2 = 0
n^2 + 2n + n + 2 = 0
n(n + 2) + 1(n + 2) = 0
(n + 1)(n + 2) = 0
n = -1 ...OR... n = -2

(n+1)(n+2)=0
n=-1 & n=-2

This is easier done with factoring, but if you have to use the quadratic formula:

Refer to my Math Variety blog post on how to solve quadratic equations by using the quadratic formula and feel free to click on any ads/links of interest
http://mathvariety.blogspot.com/2009/04/solving-quadratic-equations-by-using.html

Check out my eHow article on how to solve quadratic equations by using the quadratic formula:
http://www.ehow.com/how_4816180_quadratic-equation-using-quadratic-formula.html

Also check out this link. It’s a calculator that will give you the roots (solutions) to any quadratic equation. Just enter values for a, b, and c.
http://www.1728.com/quadratc.htm

The quadratic formula is:
n = [-b +/- sqrt(b^2-4ac)]/(2a)

Here, a=1, b=3, c=2. We have:
n=[-3 +/- sqrt(3^2-4*1*2)]/(2*1)
n=[-3 +/- sqrt(9-8)]/2
n=[-3 +/- sqrt(1)]/2
n= -1 or n= -2

n= -2 and -1

first factorize it (hope you know)
3n can be split into 2 + 1
so, 0 = (n+2)(n+1)
n + 2 =0, n +1=0
n = -2 and n= -1

n^2 + 3n + 2 = 0
n^2 + 2n + n + 2 = 0
(n^2 + 2n) + (n + 2) = 0
n(n + 2) + 1(n + 2) = 0
(n + 2)(n + 1) = 0

n + 2 = 0
n = -2

n + 1 = 0
n = -1

∴ n = -2 , -1

The easy way is factoring.
(n + 2)(n + 1) = 0
n = -1, -2

n = [-3 +/- sqrt(3^3 - 4*2)]/2
n = [-3 +/- sqrt(9 - 8)]/2
n = [-3 +/- sqrt 1]/2
n = [-3 +/- 1]/2
n = (-3 - 1)/2 = -4/2 = -2
n = (-3 + 1)/2 = -2/2 = -1

0=(n+2)(n+1)
n+2=0 n+1=0
n= -2 n= -1

No. that is not the quadratic formula. Make it clear what you are asking.