Substitute the values for a, b, and c into b2 - 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two x-intercepts.) Check all that apply.

0 = 2x2 - 7x - 9

0 = x2 - 4x + 4

0 = 4x2 - 3x - 1

0 = x2 - 2x - 8

0 = 3x2 + 5x + 3

That is when b^2-4ac>0, that is when you have 2 real solutions

alright, so ax^2+bx+c=0

first one:

a=2, b=-7, c=-9

b^2-4ac = (-7) ^2-4 (2) (-9) = 49+72>0, so this has 2 real solutions

2nd one

a=1, b=-4, c=4

b^2-4ac = (-4) ^2-4 (1) (4) = 16-16=0, so this has only 1 real number solution

3rd

a=4, b=-3, c=-1

b^2-4ac = (-3) ^2-4 (4) (-1) = 9+16>0, so this has 2 real number solutions

4th

a=1, b=-2, c=-8

b^2-4ac = (-2) ^2-4 (1) (-8) = 4+32>0, so this has 2 real number solutions

5th

a=3, b=5, c=3

b^2-4ac = (5) ^2-4 (3) (3) = 25-36<0, this has no real soltuions

answer is 1st, 3rd, 4th

those are

0 = 2x2 - 7x - 9

0 = 4x2 - 3x - 1

0 = x2 - 2x - 8