Answer:
c. zero
Step-by-step explanation:
The expression on the left of the equal sign is a polynomial of degree 2. (The highest power of x is 2.) A polynomial of degree 2 is called a "quadratic." Values of the variable (x, in this case) that make the value of the quadratic be zero are called "zeros" or "roots" of the quadratic.
Every polynomial has as many roots as its degree. So, a second degree polynomial (quadratic) will have two roots. The roots may be real numbers, or they may be complex numbers. For polynomials of degree higher than 2, there may be some roots of each kind.
This question is asking, "How many roots of this quadratic are real numbers?"
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There are several different ways you can figure out the answer to this question. One of the simplest is to graph the quadratic. (See attached.) You can see that the graph of the quadratic never has a y-value of zero, so there are no (real) values of x that will be solutions to this equation.
The two solutions are -0.25±i√1.1875. The "i" indicates that portion of the number is imaginary, and the entire number (real part plus imaginary part) is called a "complex" number. Both solutions for this quadratic are complex, not real.
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Another way you can answer this question is to compute what is called the "discriminant." The roots of every quadratic of the form ax^2+bx+c can be found using the formula ...
x = (-b±√(b^2-4ac))/(2a)
For this quadratic, the values of a, b, and c are 4, 2, and 5, respectively. Then the formula becomes ...
x = (-2±√(2^2 -4·4·5))/(2·4) = (-2±√-76)/8
The value under the radical sign is the "discriminant." When it is negative, as here, the value of the square root is an imaginary number (not a real number), so the roots are complex. When the discriminant is zero, the two roots have the same value; when it is positive, there are two distinct roots.
There are zero real number solutions to this equation.
