So what we want here is a particular solution to our ODE given our condition: $y(-4) = -4$
y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4
$\Rightarrow \dfrac{dy}{dx} = \sqrt{-2y(x) + 28}$
$\Rightarrow \int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$
$u = 28-2y$
$du = -2dx$
$dx = -\dfrac{1}{2}du$
$\Rightarrow -\dfrac{1}{2}\int {1 \over \sqrt{u} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$
$\Rightarrow -\dfrac{1}{2}\int {u^{-\dfrac{1}{2}}} \hspace{2 pt}\text{d}y = \int {1}~\text{d}x$
$\Rightarrow -\dfrac{1}{2}2u^{\dfrac{1}{2}} = x$
$\Rightarrow -\sqrt{28-2y} = x + c$
$\Rightarrow \sqrt{28-2y} = -c - x,~~y(-4) = -4$
$\Rightarrow \sqrt{28-2(-4)} = -c - (-4)$
$\Rightarrow \sqrt{36} = -c + 4$
$\Rightarrow 6 = -c + 4$
$\Rightarrow c = -2$
$\Rightarrow \sqrt{28-2y}^{2} = (-c-x)^2$
$\Rightarrow 28-2y = (-(-2)-x)$
$\Rightarrow 28-2y = (2-x)^{2}$
$\Rightarrow 28-2y = 4-4x+x^{2}$
$\Rightarrow 2y = 28-4+4x-x^{2}$
$\Rightarrow y(x) = \dfrac{28-4+4x-x^{2}}{2}$
$\Rightarrow y(x) = \dfrac{28}{2}-\dfrac{4}{2}+\dfrac{4x}{2}-\dfrac{x^{2}}{2}$
$\Rightarrow y(x) = 14-2+2x-\dfrac{1}{2}x^{2}$
$\Rightarrow y(x) = -\dfrac{1}{2}x^{2}+2x+12.$
Hence,
$~~~~~~~~~~~~~~~~~~~~~~~~y(x) = -\dfrac{1}{2}x^{2}+2x+12$
is our particular solution found to our original first-order seperable linear ordinary differential equation. $\blacksquare$
I hope this helped out, and hopefully I did not make any mistakes to cause any type of confusion
here.