Answer:
He could have 4 dimes, or 8 nickles, or 1 quater and 3 nickels, or 1 quater with one dime and 1 nickel, or 3 dimes and 2 nickles, or 1 quater with obe nickel and one dime.
Step-by-step explanation:
hope this helps.
9x²+49x=22-5x
==>9x²+54x-22=0
==>9(x²+2*3x+9)-22-81=0
==>9(x+3)²-103=0
