Answer:
1. has solution x=-2 and goes in the left circle
2. has solutions x=-2 and x=5 and goes in the intersection between the two circles
3. has solution x=-2 and goes in the left circle
4. has solution x=5 and goes in the right circle
Explanation:
Solve each equation by factoring
2x^2-12x-32=0
2(x^2-6x-16)=0
2(x+2)(x-8)
x=-2,x=8
Therefore, 1. has solution x=-2 and goes in the left circle
2x^2-6x-20=0
2 (x^2-3x-10) = 0
2(x-5)(x+2)=0
x=-2,x=5
Therefore, 2. has solutions x=-2 and x=5 and goes in the intersection between the two circles
3x^2+2x-8=0
(x+2)(3x-4)=0
x=-2, x=4/3
Therefore, 3. has solution x=-2 and goes in the left circle
3x^2-18x+15=0
3(x^2-6x+5)=0
3(x-1)(x-5)=0
x=1,x=5
Therefore 4. has solution x=5 and goes in the right circle