__The equation in the problem is:__

$x^2+10x+50=-4x+1$First, **we identify that the equation is quadratic** (and this is because the quadratic term in it does not cancel out), therefore, **we will simplify the equation by moving all terms to one side **and combine thelike terms:

$x^2+10x+50=-4x+1 \\
x^2+10x+4x+50-1 =0 \\
x^2+14x+49 =0 \\$

We want to solve this equation using factorization.

**First, we'll check if we can find a common factor,** but this is not possible, since there is no multiplicative factor common to all three terms on the left side of the equation.

**We can factor the expression on the left side using the quadratic factoring formula for a trinomial**, __however__, we prefer to factor it using the trinomial factoring method__l:__

Note that the coefficient of the quadratic term (the term with the second power) is 1, and therefore we can try to perform factoring according to the __quick trinomial method__:

But before we do this in the problem - let's remember the general rule for factoring with the** **__quick trinomial method__:

The rule states that for the algebraic quadratic expression:

$x^2+bx+c$We can find a factorization to the form of a product if we can find two numbers $m,\hspace{4pt}n$such that the conditions (__conditions of the quick trinomial method__) are met:

$\begin{cases}
m\cdot n=c\\
m+n=b
\end{cases}$**If** we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression mentioned above into the form of a product and present it as:

$x^2+bx+c \\
\downarrow\\
(x+m)(x+n)$which is its factored form (product factors) of the expression,

**Let's return** now to the equation in the problem that we received in the last stage after arranging it:

$x^2+14x+49 =0$Note that** the coefficients** from the general form we mentioned in the rule above:

$x^2+bx+c$**are:**$\begin{cases}
c=49 \\
b=14
\end{cases}$Don't forget__ to consider the coefficient together with its sign.__

Let's continue - we want to factor the expression on the left side into factors according to the __quick trinomial method__, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$ that satisfy:

$\begin{cases}
m\cdot n=49\\
m+n=14
\end{cases}$We'll try to identify this pair of numbers __using our knowledge of the multiplication table__, we'll start from the multiplication between the two required numbers $m,\hspace{4pt}n$ that is - from the first row of the pair of requirements we mentioned in the last stage:

$m\cdot n=49$We identify **that their product needs to give a positive result**, and therefore we can conclude **that their signs are identical.**

Next, we'll refer to the factors (integers) of the number 49, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 7 and 7, or 49 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities for the second condition:

$m+n=14$ will lead to a quick conclusion __that the only possibility__ for fulfilling both of the above conditions together is:

$7,\hspace{4pt}7$That is:

$m=7,\hspace{4pt}n=7$__(It doesn't matter which one we call m and which one we call n)__

It is satisfied that:

$\begin{cases}
\underline{7}\cdot \underline{7}=49\\
\underline{7}+\underline{7}=14
\end{cases}$ From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+14x+49 \\
\downarrow\\
(x+7)(x+7)$

In other words, we performed:

$x^2+bx+c \\
\downarrow\\
(x+m)(x+n)$

**If so** we factored the quadratic expression on the left side of the equation into factors using factoring according to the __quick trinomial method__, and the equation is:

$x^2+14x+49=0 \\
\downarrow\\
(x+7)(x+7)=0\\
(x+7)^2=0\\$In the last stage we notice that the expression on the left side the term:

$(x+7)$

is multiplied by itself and therefore the expression can be written as a squared term:

$(x+7)^2$

Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to the quick solution of the equation we received:

$(x+7)^2=0$

**Let's pay attention to a simple fact**, on the left side there is a term that is raised to the second power, and on the right side the number 0.

0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:

$x+7=0$(We could have solved algebraically and taken the square root of both sides of the equation, **we'll discuss this in a note at the end**)

We'll solve this equation by transferring the constant number to the other side and we'll get that** the only solution** is:

$x=-7$Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method:

$x^2+14x+49=0 \\
\downarrow\\
(x+7)(x+7)=0\\
(x+7)^2=0\\
\downarrow\\
x+7=0\\
x=-7$__Therefore, the correct answer is answer B.__

__Note:__

We could have reached the final equation by **taking the square root** of both sides of the equation, **however - taking a square root involves considering two possibilities: positive and negative** (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:

$(x+7)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\
\downarrow\\
\sqrt{(x+7)^2}=\pm\sqrt{0} \\
x+7=\pm0\\
x+7=0$

On the left side, the root (which is a half power) and the second power canceled each other out, and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and **therefore we reached the same equation we reached by logic - in the solution above**.

In a case where on the right side there's a number other than 0, we could solve **only by taking the root** and considering the two positive and negative possibilities which would then give two different possibilities for the solution.