The equation of the line passing through point $ \left(3,~-4\right) $, and point $ \left(2,~0\right) $ is:
$$ 4x+y-8=0 $$To find equation of the line passing through points $ A(x_A,y_A) $ and $ B(x_B,y_B) $, we use formula:
$$ y - y_A~=~\frac{y_B - y_A}{x_B - x_A}(x-x_A) $$In this example we have:
$$ \begin{aligned} & \left(3,~-4\right) \implies x_A = 3 ~~\text{and}~~ y_A = -4 \\[1 em] & \left(2,~0\right) \implies x_B = 2 ~~\text{and}~~ y_B = 0 \end{aligned} $$After substituting into the formula, we obtain:
$$ \begin{aligned} y - y_A~&=~\frac{y_B - y_A}{x_B - x_A}(x - x_A) \\[1 em] y - \left(-4\right)~&=~\frac{ 0 - \left(-4\right) }{ 2 - 3 } \left( x - 3 \right) \\[1 em]y + 4 ~&=~ -4 \left( x - 3 \right) \\[1 em]y + 4 ~&=~ -4x + 12 \\[1 em]y ~&=~ -4x + 8 \end{aligned} $$