first off, let's put x + 2y = -2 in slope-intercept form, so can see what its slope is,
[tex]\bf x+2y=-2\implies 2y=-2-x\implies y=\cfrac{-x-2}{2}\implies y=-\cfrac{1}{2}x-1[/tex]
so, as we can see, it has a slope of -1/2.
now, a perpendicular line to that one, will have a negative reciprocal slope,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for}\quad -\cfrac{1}{2}\\\\
negative\implies +\cfrac{1}{ 2}\qquad reciprocal\implies + \cfrac{ 2}{1}\implies 2[/tex]
so, we're really looking for a line whose slope is 2 and runs through -3,-4.
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
&&(~ -3 &,& -4~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 2
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=2[x-(-3)]
\\\\\\
y+4=2(x+3)\implies y+4=2x+6\implies y=2x+2[/tex]
