Help will give brainliest

at least three are dimes

so we have 20 coins with a total of 3.50

if the rest are all quarters then there would be 4*3+2=14 but we need 20

so we are 6 short

we must have dimes in multiples of 5

12 q + 5 d=3.50 but only 17 coins

10q +10 dimes=3.50 with 20 coins

10 q +13 dimes total

algebraic solution

d+q=23

25q+10d=380

d=23-q