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To solve a quadratic equation by completing the square, we aim to form a perfect square trinomial from the quadratic expression. This method is often used to make an equation simpler to solve, especially when it's not easily factorable.
First, move any constant term to the right side of the equation, as shown in the problem from original exercise where the equation becomes:

• \(y^2 - 5y = 6\)

Next, to complete the square, take half of the coefficient of the linear term, square it, and add and subtract it inside the equation. If the coefficient of \(y\) is \(-5\), you take half of it which is \(-\frac{5}{2}\), and then square it to get \(6.25\). Add and subtract this:

• \(y^2 - 5y + 6.25 - 6.25 = 6\)

This adjustment allows you to express the left side as a perfect square trinomial:

• \((y - \frac{5}{2})^2 - 6.25 = 6\)

Finally, move all constants to the other side again to solve for \(y\):

• \((y - \frac{5}{2})^2 = 12.25\)

The last step involves solving the square by taking square roots, yielding two solutions for \(y\). In our example, they evaluate to \(y = 6\) and \(y = -1\).

The quadratic formula is a powerful tool for finding the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). It is particularly useful when the quadratic equation is difficult to factor. The formula itself is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This expression gives us the x-values that satisfy the equation.
The beauty of this formula is its universality—it can be applied to any quadratic equation:

• First, identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation.
• Next, plug these values into the formula to evaluate the possible solutions.
• The \(\pm\) symbol means there are typically two solutions, one for addition and one for subtraction.

For example, in a quadratic equation similar to \(y^2 - 5y - 6 = 0\), we can apply the quadratic formula to find the solutions without completing the square or factoring directly. This method always works but might require more computation compared to simpler methods like factoring.

Factoring is one of the quickest ways to solve quadratic equations when they are factorable. This method involves expressing the quadratic as a product of two binomials.
Consider a basic quadratic form: \(ax^2 + bx + c\). To factor it, you look for two numbers that multiply to \(ac\) and add up to \(b\).

• Write the expression as \((x + m)(x + n)\) where \(m \cdot n = ac\) and \(m + n = b\).
  
• Split the middle term using these two numbers.

This method suits simple quadratics or can be adjusted slightly for more complex equations using techniques like grouping. Once factored, set each binomial to zero to solve for the variable, giving possible solutions.
In the case of the original problem \(y^2 - 5y - 6 = 0\), you could suspect it's factorable and factor it as \((y - 6)(y + 1) = 0\), which leads to the solutions \(y = 6\) and \(y = -1\). Each method has its own advantages, and factoring can be fast and efficient when applicable.