How do you factor quadratic equations with two variables?

Hint:Factoring reduces the higher degree equation into its linear equation. In the above given question, we need to reduce the quadratic equation into its simplest form in such a way that addition of products of the factors of first term and last should be equal to the middle one.

Complete step by step solution:
\[a{x^2} + bxy + c{y^2} + dx + ey + f = 0\] is a general way of writing quadratic equations where a, b c,d,e and f are the numbers where $a,c \ne 0$
Now we can take can example of an equation,
\[2{x^2} + 7xy - 15{y^2}\]
In the above expression,
a=2, b=7, c=-15 d=0 e=0 f=0
First step is by multiplying the term \[2{x^2}\] and the constant term -15, we get \[ - 30{x^2}\].
After this, factors of \[ - 30{x^2}\] should be calculated in such a way that their addition should be equal to 7xy.
Factors of -30 can be -3 and 10
where \[ - 3xy + 10xy = 7xy\].
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
\[
\Rightarrow 2{x^2} + 7xy - 15{y^2} = 0 \\
\Rightarrow 2{x^2} - 3xy + 10xy - 15{y^2} = 0 \\
\\
\]
Now, by grouping the first and last two terms we get common factors.
\[
\Rightarrow \left( {2x - 3y} \right) + 5y\left( {2x - 3y} \right) = 0 \\
\\
\]
Taking x common from the first group and 1 common from the second we get the above equation.
We can further solve it we get,
\[
\Rightarrow \left( {x + 5y} \right)(2x - 3y) = 0 \\
\\
\]
So here we get the above reduced form.

Note: An important thing to note is that suppose there is a trinomial expression given which has leading coefficients having perfect squares then we can bring the terms by writing it in perfect square then using the formula of ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.