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To simplify the calculations, let $x+5$ be equal to $y$ .

$\left( y-3 \right) \left( y-1 \right) \left( y+1 \right) \left( y+3 \right) +16=0$

Since set of reals is closed under addition and subtraction, we just need to find the real values of $y$ .

$\left( y-3 \right) \left( y-1 \right) \left( y+1 \right) \left( y+3 \right) +16=0\\ \left[ \left( y-1 \right) \left( y+1 \right) \right] \left[ \left( y-3 \right) \left( y+3 \right) \right] =-16\\ \left( { y }^{ 2 }-1 \right) \left( { y }^{ 2 }-9 \right) =-16\\ { y }^{ 4 }-10{ y }^{ 2 }+9=-16\\ { y }^{ 4 }-10{ y }^{ 2 }+25=0\\ { \left( { y }^{ 2 }-5 \right) }^{ 2 }=0$

We have 2 double roots. One simplifying,

${ y }^{ 2 }=5\\ y=\pm \sqrt { 5 }$

Hence the equation has exactly 2 real roots.