Instead of doing it for you, I'll do one just like it:
Find two consecutive odd integers such that their product is 1 less than 6 times their sum.
Let the smaller one be x and larger be x+2.
x(x+2) = 6[x+(x+2)]-1
x²+2x = 6[x+x+2]-1
x²+2x = 6[2x+2]-1
x²+2x = 12x+12-1
x²+2x = 12x+11
x²-10x-11 = 0
(x-11)(x+1) = 0
x-11=0; x+1=0
x=11; x=-1
One answer is 11 and 13, the other answers are -1 and +1
Checking 11 and 13. Their product is 143, which is 1 less than
6 times their sum, which is 6 times 11+13=24 and 6×24 = 144, and
143 is 1 less than 144. So 11 and 13 checks.
Checking -1 and 1. Their product is -1, which is 1 less than
6 times their sum, which is 6 times -1+1 = 0 and 6×0 = 0, and
-1 is 1 less than 0. So -1 and 1 checks.
Now use this as a guide and solve yours. This problem has a
"less than" and yours has a "more than". So you'll have to add
instead of subtract.
Edwin