History of Hindu Mathematics: A Source Book v2.c17.s21.18

v2.c17.s21.18

Substituting this value in the equation bx + c = y², we get bx + c &= (bu ± r)² / &= b²u² ± 2bru + r², / or bx + c - r² &= b²u² ± 2bru, / or bx + b(m - s) &= b²u² ± 2bru. / \therefore x &= bu² ± 2ru - (m - s).