Close-up of handwritten mathematical equations and solutions, showcasing algebraic problem-solving techniques.

Let’s solve this step by step.

We are given a quadratic equation:

5x² + 2x = 8

First, we need to rewrite it in standard form (ax² + bx + c = 0), so subtract 8 from both sides:

→ 5x² + 2x - 8 = 0

Now, we’ll use the quadratic formula to find the solutions:

> x = [-b ± √(b² - 4ac)] / (2a)

Here:
- a = 5
- b = 2
- c = -8

Step 1: Calculate the discriminant (the part under the square root):

Discriminant = b² - 4ac
= (2)² - 4*(5)*(-8)
= 4 - (-160)
= 4 + 160
= 164

Since 164 is positive but not a perfect square, we will have two real irrational roots.

Step 2: Plug into the formula:

x = [-2 ± √164] / (2*5)
x = [-2 ± √164] / 10

Simplify √164:

√164 = √(4 * 41) = 2√41

So:

x = [-2 ± 2√41] / 10

Factor out 2 in numerator:

x = 2[-1 ± √41] / 10
x = [-1 ± √41] / 5

So the two solutions are:

→ x = (-1 + √41)/5
→ x = (-1 - √41)/5

These are two real, irrational numbers — one positive and one negative (since √41 ≈ 6.4, so -1 + 6.4 = 5.4 → positive; -1 - 6.4 = -7.4 → negative).

This matches what the problem says: “Some should always be 1 positive and 1 negative solution.”

Final Answer:
x = \frac{-1 + \sqrt{41}}{5} \quad \text{and} \quad x = \frac{-1 - \sqrt{41}}{5}