The complete question is
"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 = 2x^2 – 7x – 9
0 = 4x^ 2 – 3x – 1
The quadratic equations that have real number solutions are; 4x^2 – 3x – 1, and 2x^2 – 7x – 9.
The formula for finding the discriminant is
[tex]b^2 - 4ac[/tex]
The solution contains the term [tex]\sqrt{b^2 - 4ac}[/tex] which will be:
Real and distinct if the discriminant is positive
Real and equal if the discriminant is 0
Non-real and distinct roots if the discriminant is negative
For the quadratic equation [tex]2x^2 - 7x - 9[/tex]
[tex]b^2 - 4ac[/tex]
[tex]= (-7) ^2 - 4( 2) ( -9)\\\\= 49 + 72 = 121[/tex]
This equation has two real number solutions.
For the quadratic equation [tex]4x^ 2 - 3x- 1[/tex]
[tex]b^2 - 4ac[/tex]
[tex]= (-3) ^2 - 4( 4) ( -1)\\\\= 9 + 16 = 25[/tex]
This equation will have two real number solutions.
Learn more on discriminant here:
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