Question

(x^2+6x)^2+2(x+3)^2=81

Answer 1

$(x^2+6x)^2+2(x+3)^2=81$
$(x^2+6x)^2+2(x^2+6x+9)=81$
$(x^2+6x)^2+2(x^2+6x)+18=81$
$(x^2+6x)^2+2(x^2+6x)+18-81=0$
$(x^2+6x)^2+2(x^2+6x)-63=0$
Замена:  $x^2+6x=b$
$b^2+2b-63=0$
$D=2^2-4*1*(-63)=256=16^2$
$b_1= \frac{-2+16}{2} =7$
$b_2= \frac{-2-16}{2} =-9$

$x^{2} +6x=7$                                   или                  $x^{2} +6x=-9$
$x^{2} +6x-7=0$                            или                  $x^{2} +6x+9=0$
$D=6^2-4*1*(-7)=64$          или                  $(x+3)^2=0$
$x_1= \frac{-6+8}{2} =1$                                или                  $x=-3$
$x_2= \frac{-6-8}{2} =-7$
Ответ:  $-7;$   $-3;$   $1$

Answer 2

Let's solve this quadratic equation. First, expand and simplify the equation:

$(x^2+6x)^2 + 2(x+3)^2 = 81$

$(x^2+6x)(x^2+6x) + 2(x+3)(x+3) = 81$

Now, multiply the terms:

$x^4 + 12x^3 + 36x^2 + 2(x^2 + 6x + 9) = 81$

Combine like terms:

$x^4 + 12x^3 + 36x^2 + 2x^2 + 12x + 18 = 81$

Combine further:

$x^4 + 12x^3 + 38x^2 + 12x + 18 = 81$

Now, subtract 81 from both sides:

$x^4 + 12x^3 + 38x^2 + 12x - 63 = 0$

This is a quartic equation, and solving it directly can be complex. If you need the exact solutions, you might want to use numerical methods or a graphing calculator.