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To solve quadratic equations such as \((x+2)^{2}-50=0\), one of the key steps is to isolate the square term. This involves rearranging the equation so that the term with the exponent is by itself on one side of the equation. To accomplish this, we often move any constant terms to the opposite side of the equation. For instance, by adding 50 to both sides of \((x+2)^{2}-50=0\), we isolate the square term, yielding \((x+2)^2 = 50\). This step is essential because it sets the stage for us to use the next technique - taking square roots - to find the solution for the variable.

In the context of our problem, isolating the square term is step 1, which simplifies the equation and makes it ready for the next steps in finding the solution. Students must remember that the goal is to have only the squared term and its coefficient, if any, on one side of the equation to proceed effectively with the solution process.

With the square term isolated, the next step in solving a quadratic equation like \((x+2)^2 = 50\) is taking square roots of both sides of the equation. When you apply the square root to both sides, it is crucial to remember that there will always be two solutions: one positive and one negative, because squaring either positive or negative numbers yields a positive result. Hence, the square root of a positive number is both its positive and negative values.

For our exercise, taking the square root of both sides gives us two equations: \(x + 2 = +\sqrt{50}\) and \(x + 2 = -\sqrt{50}\). This step might introduce square roots that are not perfect squares, which can lead to exact solutions that involve irrational numbers (numbers that cannot be expressed as a simple fraction), represented as a square root, or rounded decimal solutions.

When we reach the final steps of solving quadratic equations, we aim to express the solutions exactly and, if necessary, as decimal approximations. Exact solutions are representations of the solutions in the most precise form, typically involving square roots when the equation does not factor into whole numbers. In contrast, decimal solutions are approximate values, often rounded to a specified place value as per the problem's instructions.

In the given exercise, the exact solutions are \(x = \sqrt{50} - 2\) and \(x = -\sqrt{50} - 2\). These exact forms may not always be practical for real-world measurements or when comparing values, so we also provide decimal approximations. Here, when rounded to the nearest tenth, the solutions translate into \(x = 5.1\) and \(x = -9.1\). Students should be adept at finding both exact and decimal solutions to be versatile in handling different mathematical and practical situations, explicitly following rounding protocols when working with approximations.