Apr 10th, 2013 at 03:49 PM

# `ax^2+bx+c=0` what are the roots of this quadratic equation? OR Solve the quadratic equation `ax2 + bx+c=0`. #Solution: `ax2 + bx+c=0` =>`x2+b/a x+c/a=0` =>`(x)2+2(x)( b/2a) +( b/2a)2 + c/a - ( b/2a)2 =0` =>`(x+ b/2a)2 = - c/a + ( b/2a)2` =>`(x+ b/2a)2 = - c/a + b^2/(4a^2 )` =>`(x+ b/2a)2 = (b^2-4ac)/(4a^2 )` =>`x+ b/2a = ±√((b^2-4ac)/(4a^2 ))` =>`x = - b/2a ±√(b^2-4ac)/2a ` =>`x = (-b±√(b^2-4ac))/2a`