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Q: What does all real number are solutions?
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What two numbers go into 28?

1,2,4,7,14,28. These are only the whole number factors. Of course, there are an infinite number of solutions to this because all that is required it that xy=28. If x and y are unrestricted, and their domain is all real numbers, there are infinite solutions.


What are the solutions of a six-degree polynomial?

It depends on the polynomial and your degree of sophistication. In the complex domain, it will have six solutions, although not all of them need be different. If the coefficients are all real, then it will have 0, 2, 4 or 6 real solutions in the real domain.


What is the discriminant to determine how many real number solutions the quadratic equation -4j2 plus 3j-28 equals 0 has?

The discriminant is -439 and so there are no real solutions.


When are these kinds of numbers solutions to quadratic equations?

You need to be more specific. A quadratic equation will have 2 solutions. The 2 solutions can be equal (such as x&sup2; + 2x + 1 = 0, solution is -1 and -1). If one of the solutions is a real number, then the other solution will also be a real number. If one of the solutions is a complex number, then the other solution will also be a complex number. [a complex number has a real component and an imaginary component]In the equation: Ax&sup2; + Bx + C = 0. The term [B&sup2; - 4AC] will determine if the solution is a double-root, or if the answer is real or complex.if B&sup2; = 4AC, then a double-root (real).if B&sup2; > 4AC, then 2 real rootsif B&sup2; < 4AC, then the quadratic formula will produce a square root of a negative number, and the solution will be 2 complex numbers.If B = 0, then the numbers will be either pure imaginary or real, and negatives of each other [ example 2i and -2i are solutions to x&sup2; + 4 = 0]Example of 2 real and opposite sign: x&sup2; - 4 = 0; 2 and -2 are solutions.


How many real solutions can the quadratic formula give?

A quadratic equation can have either two real solutions or no real solutions.