Which of the following quadratic equations has -4 and 3/4 as the solutions?

a.)x^2+(13/4x)-3
b.)x^2+(30/7x)-(25/7)
c.)x^2+(30/7x)+(25/7)
d.)x^2-(13/4x)+3

I substituted the values and I have the answer as (a.)

To determine which of the given quadratic equations has -4 and 3/4 as the solutions, you can use the zero product property. The zero product property states that if ab = 0, then either a = 0 or b = 0.

To find the solutions of a quadratic equation, you need to set the equation equal to zero and then factor it. In this case, let's set each of the given quadratic equations equal to zero and factor them:

a.) x^2 + (13/4)x - 3 = 0
b.) x^2 + (30/7)x - (25/7) = 0
c.) x^2 + (30/7)x + (25/7) = 0
d.) x^2 - (13/4)x + 3 = 0

Now, substitute -4 and 3/4 into each equation:

a.) (-4)^2 + (13/4)(-4) - 3 = 0
b.) (3/4)^2 + (30/7)(3/4) - (25/7) = 0
c.) (3/4)^2 + (30/7)(3/4) + (25/7) = 0
d.) (-4)^2 - (13/4)(-4) + 3 = 0

By evaluating these equations, we can see that option (a) is the only one that satisfies the given solutions of -4 and 3/4. Therefore, the answer is (a.) x^2 + (13/4)x - 3.