Solve for 
\[x\]. Enter the solutions from least to greatest.
\[5x^2 + 15x - 50 = 0\]
\[\text{lesser }x = \] 
\[\text{greater } x = \]

We are given the quadratic equation:

\[
5x^2 + 15x - 50 = 0
\]

### Step 1: Simplify the equation
We can factor out 5 from all the terms to simplify:

\[
x^2 + 3x - 10 = 0
\]

### Step 2: Solve using the quadratic formula
The quadratic formula is:

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

For the equation \(x^2 + 3x - 10 = 0\), the coefficients are:
- \(a = 1\)
- \(b = 3\)
- \(c = -10\)

Plugging these values into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)}
\]

\[
x = \frac{-3 \pm \sqrt{9 + 40}}{2}
\]

\[
x = \frac{-3 \pm \sqrt{49}}{2}
\]

\[
x = \frac{-3 \pm 7}{2}
\]

### Step 3: Calculate the two solutions

- For the positive root:
\[
x = \frac{-3 + 7}{2} = \frac{4}{2} = 2
\]

- For the negative root:
\[
x = \frac{-3 - 7}{2} = \frac{-10}{2} = -5
\]

### Final solutions:

- Lesser \(x = -5\)
- Greater \(x = 2\)

Let me know if you'd like more details, or have any questions!

---

Here are 5 related questions for further exploration:
1. How would the solutions change if the constant term in the equation were different?
2. Can this equation be solved by factoring instead of using the quadratic formula?
3. What is the vertex of the parabola represented by \(5x^2 + 15x - 50 = 0\)?
4. How can you verify the solutions by plugging them back into the original equation?
5. How does completing the square compare with the quadratic formula for this equation?

**Tip**: Always check if a quadratic equation can be simplified before solving. Factoring common terms can make the calculation easier!

Solve for x. Enter the solutions from least to greatest.
5x^2 + 15x - 50 = 0
Lesser x = 
Greater x = 

Learn how to solve the quadratic equation 5x^2 + 15x - 50 using the quadratic formula. This detailed step-by-step solution simplifies the equation and applies the quadratic formula to find the roots x = -5 and x = 2. Suitable for students in Grades 9-10.

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