For this, we have to draw the graph first. As
#x^2+3x+2=x^2+2xx3/2xx x+(3/2)^2+2-9/4=(x+3/2)^2-1/4#
As #(x+3/2)^2# is positive and its minimum value is zero, minimum value of #x^2+3x+2# is #-1/4#, at #x=-3/2#.
Let #y=x^2+3x+2# and we have a minimum at #(-3/2,-1/4)#, some other points, we have by putting various values of #x# (around #x=-3/2#), are
#(-5,12), (-4,6), (-3,2), (-2,0), (-1,0), (0,2), (1,6), (2,12)# and joining all these eight points and the minimum at #(-3/2,-1/4)#, we get following graph.
graph{x^2+3x+2 [-4.333, 0.667, -0.43, 2.07]}
As the value of #y=0# at #x=-1# and #x=-2#,
the solution is #x=-1# and #x=-2#.
