Answer:
[tex]\textsf{D.} \quad x=4 \pm \sqrt{17}[/tex]
Step-by-step explanation:
Rearrange the equation so it equals zero:
[tex]\begin{aligned} x^2+4 & =8x+5 \\ \implies x^2+4-8x-5 & = 0\\x^2-8x-1&=0\end{aligned}[/tex]
Now use the quadratic formula to solve for x:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
[tex]a=1, \quad b=-8, \quad c=-1[/tex]
[tex]\implies x=\dfrac{-(-8) \pm \sqrt{(-8)^2-4(1)(-1)}}{2(1)}[/tex]
[tex]\implies x=\dfrac{8 \pm \sqrt{68}}{2}[/tex]
[tex]\implies x=\dfrac{8 \pm \sqrt{4\cdot 17}}{2}[/tex]
[tex]\implies x=\dfrac{8 \pm \sqrt{4}\sqrt{17}}{2}[/tex]
[tex]\implies x=\dfrac{8 \pm 2\sqrt{17}}{2}[/tex]
[tex]\implies x=4 \pm \sqrt{17}[/tex]
