Find two POSITIVE consecutive ODD integers such that 3 times the square of the larger is 45 more than 10 times the smaller

Answer 1

3L^2 = 45+10S

L=S+2

3(S+2)^2 = 45 + 10S

3S^2 + 12S + 12 = 45 + 10S

3S^2 +2S - 33 = 0

(3S +11)(S-3) = 0

S=3 = the smaller odd integer

L =5 = the larger odd integer

3(5)^2 = 45+ 10(3)

75 = 75