If one of the lines given by $6x^2-xy+4c{y}^2=0$ is $3x+4y=0,$ then $c$ equals

If $3x+4y=0$ is one of the lines, its slope $m_1=-3/4$.
Let the equation be $6x^2-xy+4c{y}^2=0.$ Dividing by $x^2$ gives a quadratic in $m$ ($m=y/x$):
$4c{m}^2-m+6=0$.
Since $m_1=-3/4$ is a root of this equation, substitute it:
$4c(-3/4{)}^2-(-3/4)+6=0$
$4c(9/16)+3/4+6=0$
$\frac{9c}{4}+\frac{3}{4}+\frac{24}{4}=0$
$9c+3+24=0$
$9c+27=0$
$9c=-27\Rightarrow c=-3$.