Solve the Quadratic Equation: -x² + 13x - 14 = 0

Solve the following quadratic equation:

$-x^2+13x-14=0$

Solve the following equation:

$-x^2+13x+14=0$

Let's begin by arranging the equation, making sure that the coefficient of the quadratic term is positive, we'll do this by multiplying both sides of the equation by $(-1)$:

$-x^2+13x+14=0 \hspace{6pt}\text{/}\cdot(-1) \\ x^2-13x-14=0$

Now we notice that the coefficient of the quadratic 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:

$m\cdot n=-14\\ m+n=-13\\$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 the multiplication rules. Possible factors of 14 are 2 and 7 or 14 and 1, fulfilling the second requirement mentioned. Furthermore the signs of the numbers we're looking for are different from each other leading us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-14\\ n=1 \end{cases}$

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

$x^2-13x-14=0 \\ \downarrow\\ (x-14)(x+1)=0$

From here 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 to solve them by isolating the unknown in each:

$x-14=0\\ \boxed{x=14}$

or:

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

Let's summarize the solution of the equation:

$-x^2+13x+14=0 \hspace{6pt}\text{/}\cdot(-1) \\ x^2-13x-14=0\\ \downarrow\\ (x-14)(x+1)=0 \\ \downarrow\\ x-14=0\rightarrow\boxed{x=14}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=14,-1}$

Therefore the correct answer is answer B.