Найдите сумму корней уравнения

$4( {x}^{2} + \frac{1}{ {x}^{2} } ) + 12(x + \frac{1}{x} ) - 47 = 0$
$x + \frac{1}{ {x} } = y \\ {x}^{2} + \frac{1}{ {x}^{2} } = {y}^{2} - 2$
4(у^2-2)+12у-47=0
4у^2-8+12у-47=0
4у^2+12у-55=0
Д=144+880=1024=32^2
у=-12-32/8=-11/2
у=-12+32/8=5/2

$x + \frac{1}{x} = - \frac{11}{2} \\ 2 {x }^{2} + 11x + 2 = 0 \\ d = 121 - 16 = 105 \\ x = \frac{ - 11 + - \sqrt{105} }{4}$
$x + \frac{1}{x} - \frac{5}{2 } = 0 \\ 2 {x}^{2} - 5x + 2 = 0 \\ x = 2 \\ x = \frac{1}{2}$

х1+х2=-в/а=-11/2
х3+х4=-в/а=5/2
-11/2+5/2=-3