At a high school Algebra 2 level how do you solve x^2+4x +3 step by step?

We are asked to "solve" x^2+4x+3:

In the typical Algebra assignment you are asked to either evaluate an expression, simplify and expression, or solve an equation. Each of these, while related, involve a different approach.

We could evaluate x^2+4x+3 for a given value of x; say if x=3 then the expression has the value 3^2+4(3)+3=24.

If we are asked to simplify an expression, we strive to remove grouping symbols (parantheses, brackets, etc...) and then add/subtract like terms.

Here we are asked to "solve" x^2+4x+3. It is implied that we are to solve the general equation x^2+4x+3=0.

(a) You can factor the polynomial:

(x+3)(x+1)=0  

** One method is to rewrite the linear term as the sum of two terms where the product of the coefficients is the constant term: x^2+3x+1x+3; then factor by grouping: x(x+3)+1(x+3); use the distributive property to rewrite as (x+3)(x+1)=0 **

Now use the zero product property (if ab=0 then a=0, b=0, or a=b=0) to get:

x+3=0 ==> x=-3   x+1=0 ==> x=-1

So the solutions are x=-1 or -3

(b) You could complete the square:

x^2+4x=-3
x^2+4x+4=-3+4
(x+2)^2=1

x+2=1 or x+2=-1 

x=-1 or x=-3

(c) You could use the quadratic formula

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The solutions are x=-1 or x=-3