user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x20(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L25
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jne    L25
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L35
L25:
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm2
  movsd  D7(%rip),%xmm1
  mulsd  %xmm1,%xmm2
  movsd  -0x10(%rbp),%xmm1
  mulsd  %xmm1,%xmm2
  subsd  %xmm2,%xmm0
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  jbe    L26
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L35
L26:
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm2
  movsd  D7(%rip),%xmm1
  mulsd  %xmm1,%xmm2
  movsd  -0x10(%rbp),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movsd  -0x18(%rbp),%xmm1
  movq   %rax,%xmm0
  subsd  %xmm1,%xmm0
  movsd  D99(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D105(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  -0x18(%rbp),%xmm0
  movq   D78(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm2
  movsd  D7(%rip),%xmm1
  mulsd  %xmm1,%xmm2
  movsd  -0x10(%rbp),%xmm1
  mulsd  %xmm2,%xmm1
  movapd %xmm0,%xmm3
  subsd  %xmm1,%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  movsd  -0x28(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  D99(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L35:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D60
data_type: string
name: retrieve_data
arguments: data_tag: D24
data_type: string
name: retrieve_data
arguments: data_tag: D7
data_type: byte[8]
name: retrieve_data
arguments: data_tag: D99
data_type: byte[8]
name: retrieve_data
arguments: data_tag: D105
data_type: string
name: retrieve_data
arguments: data_tag: D78
data_type: byte[8]
name: retrieve_data
arguments: data_tag: D37
data_type: string

tool: D60: %lf %lf %lf

tool: D24: Impossivel calcular

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: {"D99": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

tool: D105: R1 = %.5lf


tool: {"D78": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: D37: R2 = %.5lf


assistant: ```c
int main() {
  double a, b, c;
  scanf("%lf %lf %lf", &a, &b, &c);
  if (a == 0)
    printf("Impossivel calcular\n");
  else if (b * b - 4 * a * c < 0)
    printf("Impossivel calcular\n");
  else {
    printf("R1 = %.5lf\n", (-b + sqrt(b * b - 4 * a * c)) / 2 / a);
    printf("R2 = %.5lf\n", (-b - sqrt(b * b - 4 * a * c)) / 2 / a);
  }
  return 0;
}
```