Now enter those coefficients into the quadratic formula below: $$ x=\frac{-6 \pm \sqrt{(6))^{2}-4 \cdot 5} \cdot \sqrt{5}}{2 \cdot \sqrt{10}} $$

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Let's first clarify the quadratic formula and the coefficients involved. The standard form of a quadratic equation is:

$$
ax^2 + bx + c = 0
$$

The quadratic formula to find the roots of the equation is given by:

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

From your expression, it seems like you have the following coefficients:

- $$ a = 5 $$
- $$ b = 6 $$
- $$ c = 0 $$ (since it is not explicitly mentioned, we assume it to be 0)

Now, let's substitute these coefficients into the quadratic formula step by step.

1. Calculate $$ b^2 - 4ac $$:
   $$
   b^2 = 6^2 = 36
   $$
   $$
   4ac = 4 \cdot 5 \cdot 0 = 0
   $$
   Therefore,
   $$
   b^2 - 4ac = 36 - 0 = 36
   $$

2. Now substitute into the quadratic formula:
   $$
   x = \frac{-6 \pm \sqrt{36}}{2 \cdot 5}
   $$

3. Calculate $$ \sqrt{36} $$:
   $$
   \sqrt{36} = 6
   $$

4. Substitute back into the formula:
   $$
   x = \frac{-6 \pm 6}{10}
   $$

5. This gives us two possible solutions:
   - For the positive case:
   $$
   x_1 = \frac{-6 + 6}{10} = \frac{0}{10} = 0
   $$
   - For the negative case:
   $$
   x_2 = \frac{-6 - 6}{10} = \frac{-12}{10} = -1.2
   $$

Thus, the solutions to the quadratic equation are:

$$
x = 0 \quad \text{and} \quad x = -1.2
$$
The solutions are $$ x = 0 $$ and $$ x = -1.2 $$.

Now enter those coefficients into the quadratic formula below: \( x=\frac{-6 \pm \sqrt{(6))^{2}-4 \cdot 5} \cdot \sqrt{5}}{2 \cdot \sqrt{10}} \)

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