Solve the Quadratic Equation: 2x² - 16x = -30

Let's solve the given equation:

$2x^2-16x=-30$

First, let's simplify the equation, noting that all coefficients as well as the free term are multiples of 2. Therefore we'll divide both sides of the equation by 2, then we'll proceed to rearrange it by moving terms:

$2x^2-16x=-30 \hspace{6pt}\text{/}:2 \\ x^2-8x=-15\\ x^2-8x+15=0$

Now we notice that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

We'll look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=15\\ m+n=-8\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same signs, according to multiplication rules. Possible factors of 15 are 5 and 3 or 15 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have equal signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases}m=-5 \\ n=-3\end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2-8x+15=0 \\ \downarrow\\ (x-5)(x-3)=0$

Remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain two simple equations and proceed solve them by isolating the variable in each of them:

$x-5=0\\ \boxed{x=5}$

or:

$x-3=0\\ \boxed{x=3}$

Let's summarize then the solution of the equation:

$2x^2-16x=-30\\ x^2-8x+15=0 \\ \downarrow\\ (x-5)(x-3)=0 \\ \downarrow\\ x-5=0\rightarrow\boxed{x=5}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=5,3}$

Therefore the correct answer is answer C.