Question 88446
Well, you can always use the quadratic formula to solve a quadratic equation: {{{x = (-b+-sqrt(b^2-4ac))/2a}}}
But just to make it a little less confusing, we'll write your equation like this, with a B instead of a b:
{{{2x^2+3Bx+B^2 = 0}}} so...a = 2, b = 3B, and c = B^2
Now, making the appropriate substitutions, we get:
{{{x = (-3B+-sqrt((3B)^2-4(2)(B^2)))/2(2)}}} Simplifying this, we have:
{{{x = (-3B+-sqrt(9B^2-8B^2))/4}}}
{{{x = (-3B+-sqrt(b^2))/4}}}
{{{x = (-3B+-B)/4}}}
So the roots are:
{{{x = (-3B+B)/4}}} or {{{x = -3B-B)/4}}}
{{{x = -2B/4}}} or {{{x = -4B/4}}}
{{{x = -B/2}}} or {{{x = -B}}}
But remember that we substituted the b in the original equation with B, so the answers are:
{{{x = -b/2}}} or {{{x = -b}}}