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Understanding the distributive property is crucial in algebra, especially when solving quadratic equations. This property states that multiplying a single term by a sum or difference (like the terms inside parentheses) is the same as doing each multiplication separately. In the exercise, we have the quadratic equation \(3(x^{2}+5)=4(x^{2}+2)\). To simplify, we distribute the coefficients, 3 and 4, across their respective parentheses. That is, we multiply 3 by both \(x^{2}\) and 5, and similarly, we multiply 4 by \(x^{2}\) and 2. This process transforms our equation into \(3x^2 + 15 = 4x^2 + 8\).

By systematically applying the distributive property, we've taken the first step in solving the equation by removing the parentheses, which makes combining like terms and ultimately solving the equation possible.

Once we've used the distributive property, we often need to combine like terms to simplify the equation further. Like terms are terms in an equation that have the same variable raised to the same power. In our problem, after distributive multiplication, we get \(3x^2 + 15 = 4x^2 + 8\). To proceed, we need to get all \(x^2\) terms on one side and all constant terms on the other. This is accomplished by subtracting \(3x^2\) from both sides which gives us \(x^2 = 7\). We have now effectively combined the like terms, which are the \(x^2\) terms, and we're left with a more straightforward equation to solve. Combining like terms is a way of organizing an equation, making it easier to solve for the unknown variable.

After applying the distributive property and combining like terms, we are left with the simplified equation \(x^2 = 7\). To find the value of \(x\), we now have to deal with square roots. A square root answers the question 'What number, when multiplied by itself, gives the original number?' It's important to remember that every positive number has two square roots: one positive and one negative. In our case, taking the square root of both sides of the equation \(x^2 = 7\) gives us two solutions: \(x = \sqrt{7}\) and also \(x = -\sqrt{7}\). Both of these solutions are valid because if you square both \(\sqrt{7}\) and \(–\sqrt{7}\), you’ll get 7. Square roots are fundamental in solving quadratic equations because they allow us to find the variable's value when it's raised to the power of two.