Solve 3x² + 44x - 50 = -1 - 4x² + 2x: Division-Free Method

Let's solve the given equation:

$3x^2+44x-50=-1-4x^2+2x$

$$First, let's organize the equation by moving terms and combining like terms:

$3x^2+44x-50=-1-4x^2+2x \\ 3x^2+44x-50+1+4x^2-2x=0 \\ 7x^2+42x-49=0$

Now, instead of dividing both sides of the equation by the common factor of all terms in the equation (which is 7), we'll choose to factor it out of the parentheses:

$7x^2+42x-49=0 \\ 7(x^2+6x-7)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplying expressions equals zero,

However, the first factor in the expression we got is the number 7, which is obviously different from zero, therefore:

$x^2+6x-7 =0$

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

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

$m\cdot n=-7\\ m+n=6\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to be negative, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 7 are 7 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=7\\ n=-1 \end{cases}$

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

$x^2+6x-7 =0 \\ \downarrow\\ (x+7)(x-1)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplying expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the variable on one side:

$x+7=0\\ \boxed{x=-7}$

or:

$x-1=0\\ \boxed{x=1}$

Let's summarize the solution of the equation:

$3x^2+44x-50=-1-4x^2+2x \\ 7x^2+42x-49=0 \\ \downarrow\\ 7(x^2+6x-7)=0 \\ \downarrow\\ x^2+6x-7=0\\ \downarrow\\ (x+7)(x-1)=0 \\ \downarrow\\ x+7=0\rightarrow\boxed{x=-7}\\ x-1=0\rightarrow\boxed{x=1}\\ \downarrow\\ \boxed{x=-7,1}$

Therefore the correct answer is answer D.