Factor $6x^2​ −7x−5=0$ I'm trying to factor
$$6x^2​ −7x−5=0$$
but I have no clue about how to do it. I would be able to factor this:
$$x^2-14x+40=0$$
$$a+b=-14$$
$$ab=40$$ 
But $6x^2​ −7x−5=0$ looks like it's not following the rules because of the coefficient of $x$. Any hints?

Consider the product 
\begin{align*}
(4x + 7)(3x - 8) & = 4x(3x - 8) + 7(3x - 8)\\
                 & = \color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}\\
                 & = \color{blue}{12}x^2 \color{green}{- 11}x \color{blue}{- 56}
\end{align*}
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients in the expression $\color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}$, that is,
$$(\color{blue}{12})(\color{blue}{-56}) = (\color{green}{-32})(\color{green}{21}) = -672$$
Now suppose that $ax^2 + bx + c$ is a quadratic polynomial with integer coefficients that has factorization
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
              & = rx(tx + u) + s(tx + u)\\
              & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\
              & = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su}\\
\end{align*}
Equating coefficients yields
\begin{align*}
a & = \color{blue}{rt}\\
b & = \color{green}{ru + st}\\
c & = \color{blue}{su}
\end{align*}
Note that the product of the quadratic and linear coefficients is equal to the product of the two linear coefficients that have sum $b$, that is, 
$$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st})$$
Hence, if a quadratic polynomial $ax^2 + bx + c$ factors with respect to the rational numbers, we can split the linear term into two linear terms whose coefficients have product $ac$ and sum $b$. 
To split the linear term of $6x^2 - 7x - 5$, we must find two numbers with product $6 \cdot (-5) = -30$ and sum $-7$.  They are $-10$ and $3$.  Hence,
\begin{align*}
6x^2 - 7x - 5 & = 0\\
6x^2 - 10x + 3x - 5 & = 0 && \text{split the linear term}\\
2x(3x - 5) + 1(3x - 5) & = 0 && \text{factor by grouping}\\
(2x + 1)(3x - 5) & = 0 && \text{extract the common factor}
\end{align*}
To solve the equation, use the zero product property $ab = 0 \iff a = 0~\text{or}~b = 0$.
