Who can get these correct for me ?

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

Step-by-step 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

answer:

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step-by-step explanation: