user: 这段汇编代码用C语言表示是什么样的？

```asm
<solve_equation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D84(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D95(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  -0x20(%rbp),%xmm2
  movss  D29(%rip),%xmm1
  mulss  %xmm1,%xmm2
  movss  -0x18(%rbp),%xmm1
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  comiss -0x14(%rbp),%xmm0
  jbe    L16
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L21
L16:
  movss  -0x1c(%rbp),%xmm1
  movss  -0x14(%rbp),%xmm0
  subss  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  movss  -0x20(%rbp),%xmm0
  cvtss2sd %xmm0,%xmm0
  addsd  %xmm0,%xmm0
  divsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsd2ss %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  -0x1c(%rbp),%xmm0
  movss  D6(%rip),%xmm1
  xorps  %xmm1,%xmm0
  subss  -0x14(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  movss  -0x20(%rbp),%xmm0
  cvtss2sd %xmm0,%xmm0
  addsd  %xmm0,%xmm0
  divsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsd2ss %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0xc(%rbp),%xmm0
  pxor   %xmm3,%xmm3
  cvtss2sd -0x10(%rbp),%xmm3
  movq   %xmm3,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
L21:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D84
data_type: string
name: read_data
arguments: label: D27
data_type: string
name: read_data
arguments: label: D95
data_type: string
name: read_data
arguments: label: D81
data_type: string
name: read_data
arguments: label: D29
data_type: byte[4]
name: read_data
arguments: label: D30
data_type: string
name: read_data
arguments: label: D6
data_type: byte[4]
name: read_data
arguments: label: D38
data_type: string

tool: D84: Introduce a: 

tool: D27: %f

tool: D95: Introduce b: 

tool: D81: Introduce c: 

tool: {"D29": [0x00,0x00,0x80,0x40]}

tool: D30: El resultado no entra dentro de los reales

tool: {"D6": [0x00,0x00,0x00,0x80]}

tool: D38: x1: %.3f, x2: %.3f


assistant: ```c
void solve_equation() {
  float a, b, c;
  float raiz, x1, x2;
  printf("Introduce a: ");
  scanf("%f", &a);
  printf("Introduce b: ");
  scanf("%f", &b);
  printf("Introduce c: ");
  scanf("%f", &c);
  raiz = pow(b, 2) - (4 * a * c);
  if (raiz < 0) {
    printf("El resultado no entra dentro de los reales\n");
    return;
  } else {
    x1 = (-b + raiz) / (2. * a);
    x2 = (-b - raiz) / (2. * a);
    printf("x1: %.3f, x2: %.3f\n", x1, x2);
  }
}
```