Subtract x² from both sides: Equate the coefficients of x on each side Equate the constant terms on each side: So we have this system of two equations in two unknowns: Solve the first equation for k k = 2A-5 Substitute in the second: 5(2A-5)+1 = A² 10A - 25 + 1 = A² 0 = A² - 10A + 24 0 = (A - 6)(A - 4) A - 6 = 0; A - 4 = 0 A = 6; A = 4 k = 2A-5; k = 2A-5 k = 2(6)-5; k = 2(4)-5 k = 12-5; k = 8-5 k = 3 k = 7 Edwin