History of Hindu Mathematics: A Source Book v2.c17.s8.71

i.e., \ &ax²+bx=c,\\ or \ &a²x²+abx+(b/2)²=ac+(b/2)².\\ Therefore &ax+b/2=√(ac+(b/2)²).\\ Hence \ &x=(√(ac+(b/2)²)-b/2)/a. \end{aligned}