History of Hindu Mathematics: A Source Book v2.c17.s17.t3.41

Assuming $x = √(b/ac) vy + u,$ we have $ax² + by² - c = b/c v²y² + 2√(ab/c) uvy + by²$ $+ (au² - c)$ $= b/c (au² - c) y² + 2√(ab/c) uvy + by² + v²$