Question

(x^2-7)^2-10(x^2-7)+9=0

Answer 1

$( x^{2} -7)^2-10( x^{2} -7)+9=0$
введем замену: $x^{2} -7=t$

$t^2-10t+9=0$
$D=100-36=64$
$t_1=9$
$t_2=1$

$x^{2} -7=9$   или   $x^{2} -7=1$
$x^{2} -16=0$ или   $x^{2} -8=0$
$(x-4)(x+4)=0$  или $(x+2 \sqrt{2} )(x-2 \sqrt{2} )=0$
$x=4$ или $x=-4$ или $x=-2 \sqrt{2}$ или $x=2 \sqrt{2}$
Ответ:  $4;-4;2 \sqrt{2} ;-2 \sqrt{2}$

Answer 2

$( x^{2} -7) ^{2} -10( x^{2} 7)+9=0\\ x^{2} -7=t\\ \\ t ^{2} -10t+9=0\\D=100-369=64 \\ \sqrt{D} =8 \\ t _{1} = \frac{10+8}{2} = \frac{18}{2} =9 \\ \\ t_{2} = \frac{10-8}{2} = \frac{2}{2} =1 \\ \\ x^{2} -7=9\\ x^{2} =9+7\\ x^{2} =16\\x=+-4$
$x^{2} -7=1\\ x^{2} =1+7\\ x^{2} =8\\x=+- \sqrt{8} \\ \\ x=+-2 \sqrt{2}$


$Otviet: +-4;+-2 \sqrt{2}$