Let `f(x)=x^(2)+10x+20`. Find the number of real solution of the equation

Answer

To find the number of real solutions of the equation \( f(x) = x^2 + 10x + 20 \), we will use the discriminant method. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[ D = b^2 - 4ac \]

### Step 1: Identify the coefficients
In our function \( f(x) = x^2 + 10x + 20 \), we can identify the coefficients:
- \( a = 1 \)
- \( b = 10 \)
- \( c = 20 \)

### Step 2: Calculate the discriminant
Now, we will calculate the discriminant \( D \):

\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):

\[ D = 10^2 - 4 \cdot 1 \cdot 20 \]

Calculating \( 10^2 \):

\[ D = 100 - 4 \cdot 20 \]

Calculating \( 4 \cdot 20 \):

\[ D = 100 - 80 \]

Now, subtracting:

\[ D = 20 \]

### Step 3: Determine the nature of the roots
Now that we have calculated the discriminant \( D = 20 \), we can determine the nature of the roots:
- If \( D > 0 \), the equation has two distinct real roots.
- If \( D = 0 \), the equation has exactly one real root (a repeated root).
- If \( D < 0 \), the equation has no real roots (the roots are imaginary).

Since \( D = 20 \) which is greater than 0, we conclude that:

The equation \( f(x) = x^2 + 10x + 20 \) has **two distinct real solutions**.

### Final Answer:
The number of real solutions of the equation is **2**.

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