Consider the quadratic equation x? = 4x - 5. How many solutions does the equation have? A The equation has one real solution. The equation has two real solutions. The equation has no real solutions. D The number of solutions cannot be determined​

1 real solution

{-1, 5} (two real solutions).

Step-by-step explanation:

On the left side we want x^2, not x?

We need to rewrite x? = 4x - 5 in standard form, that is, as a quadratic with x terms in decreasing powers of x:

x^2 - 4x + 5 = 0

This factors easily to (x - 5)(x + 1) = 0.

We let each factor equal zero (0) separately and solve for x:  

{-1, 5} (two real solutions).

-14 is the answer

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