Solve Quadratic Equation: Finding X in 2x²+4x-6=0

Let's solve the given equation:

$2x^2+4x-6=0$

Let's first simplify the equation, noting that all coefficients as well as the free term are multiples of the number 2, hence we'll divide both sides of the equation by 2:

$2x^2+4x-6=0 \hspace{6pt}\text{/}:2 \\ x^2+2x-3=0\\$Note 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:

$m\cdot n=-3\\ m+n=2\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules. The possible factors of 3 are 3 and 1. Fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases}m=3 \\ n=-1\end{cases}$

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

$x^2+2x-3=0\\ \downarrow\\ (x+3)(x-1)=0$

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

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

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

or:

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

Let's summarize the solution of the equation:

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

Therefore the correct answer is answer B.