Question 65841
Begin with ax^2 + bx + c = 0 and derive the quadratic formula by completing the square
{{{ax^2+bx+c=0}}}
{{{ax^2/a+bx/a+c/a=0/a}}}
{{{x^2+(b/a)x+c/a=0}}}
{{{x^2+(b/a)x+c/a-c/a=0-c/a}}}
{{{x^2+(b/a)x=-c/a}}}
{{{x^2+(b/a)x+(b/2a)^2=(b/2a)^2-(c/a)}}}
{{{(x+b/2a)^2=b^2/4a^2-c/a}}}
{{{(x+b/2a)^2=b^2/4a^2-4ac/4a^2}}}
{{{(x+b/2a)^2=(b^2-4ac)/4a^2}}}
{{{sqrt(x+b/2a)^2)=+-sqrt(((b^2-4ac)/4a^2))}}}
{{{x+b/2a=+-sqrt(b^2-4ac)/2a}}}
{{{x+b/2a-b/2a=-b/2a+-sqrt(b^2-4ac)/2a}}}
{{{highlight(x=(-b+-sqrt(b^2-4ac))/2a)}}}

then solve x = -2 - 3x^2 by using the quadratic formula
{{{3x^2+x+2=-2+2-3x^2+3x^2}}}
{{{3x^2+x+2=0}}}
a=3, b=1, c=2
{{{x=(-1+-sqrt((1)^2-4(3)(2)))/(2(3))}}}
{{{x=(-1+-sqrt(1-24))/6}}}
{{{x=(-1+-sqrt(-23))/6}}}  If you aren't doing complex numbers yet, you can stop here and say that there is no real number solution.  If you are, remember that the sqrt(-1)=i:
{{{highlight(x=(-1+-sqrt(23)i)/6)}}}
Happy Calculating!!!