algebra
posted by disha .
Solve the quadratic equation (a+b2c)x² + (2abc)x + (c+a2b) =0 is

B = 2a+b+c
B^2 = 4a^24ab4ac+b^2+2bc+c^2
4AC = 4a^2+4ab+4ac+8b^220bc+8c^2
so
B^24AC=9b^218bc+9c^2 = 9(b^22bc+c^2)
=9(bc)^2 whew!
sqrt(B^24AC) = 3(bc) = 3b  3c
so
B+/sqrt(B^24AC) = 2a+b+c +/ (3b3c)
= 2a+4b2c or 2a2b+4c
divide those by 2A