Use the quadratic formula to find the real​ solutions, if​ any, of the equation.  x^2 +8x+10=0

The quadratic equation is:

\[
x^2 + 8x + 10 = 0
\]

We will use the quadratic formula, which is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(x^2 + 8x + 10 = 0\), we can identify the values of \(a\), \(b\), and \(c\) as follows:
- \(a = 1\) (coefficient of \(x^2\))
- \(b = 8\) (coefficient of \(x\))
- \(c = 10\) (constant term)

### Step 1: Calculate the discriminant
The discriminant is given by \(b^2 - 4ac\):

\[
\text{Discriminant} = 8^2 - 4(1)(10) = 64 - 40 = 24
\]

### Step 2: Determine the nature of the roots
Since the discriminant (\(24\)) is positive, there will be two distinct real solutions, but they will involve square roots of non-perfect squares (irrational numbers).

### Step 3: Use the quadratic formula
Now we substitute the values into the quadratic formula:

\[
x = \frac{-8 \pm \sqrt{24}}{2(1)} = \frac{-8 \pm \sqrt{24}}{2}
\]

We know that \(\sqrt{24} = 2\sqrt{6}\), so:

\[
x = \frac{-8 \pm 2\sqrt{6}}{2}
\]

Simplifying:

\[
x = \frac{-8}{2} \pm \frac{2\sqrt{6}}{2} = -4 \pm \sqrt{6}
\]

### Final Answer:
The two real solutions are:

\[
x = -4 + \sqrt{6} \quad \text{and} \quad x = -4 - \sqrt{6}
\]

Would you like further details on any of the steps or concepts involved?

Here are 5 related questions to consider:
1. What is the significance of the discriminant in the quadratic formula?
2. How do irrational numbers affect the nature of the roots?
3. Can the quadratic formula be used for all types of quadratic equations?
4. What happens when the discriminant is zero?
5. How would the solutions change if the equation had a negative discriminant?

**Tip:** Always check the discriminant first to quickly understand the nature of the solutions (real or complex).

Use the quadratic formula to find the real solutions, if any, of the equation. x^2 + 8x + 10 = 0

Learn how to solve the quadratic equation x^2 + 8x + 10 = 0 using the quadratic formula. This detailed solution explains the steps to calculate the discriminant and find real solutions. The problem covers algebraic concepts and is ideal for students in Grades 8-10.

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