diff --git a/2023/puzzle-06-01.cc b/2023/puzzle-06-01.cc
new file mode 100644
index 0000000..fc2b896
--- /dev/null
+++ b/2023/puzzle-06-01.cc
@@ -0,0 +1,85 @@
+#include <cassert>
+#include <cmath>
+#include <cstdint>
+#include <cstdlib>
+#include <iostream>
+#include <iterator>
+#include <string>
+#include <string_view>
+#include <vector>
+
+using UInt = std::uint64_t;
+using NumVec = std::vector<UInt>;
+
+auto operator<<(std::ostream &os, NumVec const &vec) -> std::ostream & {
+  for (auto const num: vec) { os << ' ' << num; }
+  return os;
+}
+
+template<typename It>
+void read_nums(It it, std::string_view const nums) {
+  char const *pos{nums.data()};
+  while (pos - nums.data() < nums.size()) {
+    char *next_pos{nullptr};
+    *it++ = std::strtoull(pos, &next_pos, 10);
+    pos = next_pos;
+  }
+}
+
+[[nodiscard]] auto time_range(UInt time, UInt distance) -> UInt {
+  /* We hold down for `x` units of time and then travel (`t` - `x`) * `x` units of distance.
+   * So we want (t - x)x > d => tx - x² > d         (expand)
+   *                         => 0 > x² - tx + d     (shuffle)
+   *
+   * Lets solve the quadratic: x² - tx + d = 0
+   * with x = (-b ± √(b² - 4ac))/2a where a = 1, b = -t, c = d
+   *   => x = (t ± √(t² - 4d))/2
+   *   call two solutions x1 (negative root), x2 (positive root).
+   *
+   * Answer will the whole number of seconds between x1 and x2.  So:
+   *    floor(x2) - ceil(x1) + 1.
+   *
+   * However if x1 or x2 are the zero point we have to shift them too, so be careful of that.
+   */
+  double const t{static_cast<double>(time)};
+  double const d{static_cast<double>(distance)};
+  double const discriminant{std::sqrt(t * t - 4 * d)};
+  double const x1{(t - discriminant) / 2};
+  double const x2{(t + discriminant) / 2};
+  UInt x1i(std::ceil(x1));
+  UInt x2i(std::floor(x2));
+  if (x1i * x1i - time * x1i + distance <= 0) { ++x1i; }
+  if (x2i * x2i - time * x2i + distance <= 0) { --x2i; }
+  return x2i - x1i + 1;
+}
+
+auto main() -> int try {
+  std::string line;
+
+  NumVec times;
+  std::getline(std::cin, line);
+  assert(line.substr(0, 5) == "Time:");
+  read_nums(std::back_inserter(times), line.substr(5));
+
+  NumVec distances;
+  std::getline(std::cin, line);
+  assert(line.substr(0, 9) == "Distance:");
+  read_nums(std::back_inserter(distances), line.substr(9));
+
+  assert(distances.size() == times.size());
+
+  UInt score{1};
+  for (std::size_t i{0}; i != times.size(); ++i) {
+    auto const tr = time_range(times[i], distances[i]);
+    std::cout << times[i] << ' ' << distances[i] << " = " << tr << '\n';
+    score *= tr;
+  }
+
+  std::cout << "Score: " << score;
+
+  return EXIT_SUCCESS;
+}
+catch (...) {
+  std::cerr << "Uncaught exception.\n";
+  return EXIT_FAILURE;
+}
diff --git a/2023/puzzle-06-02.cc b/2023/puzzle-06-02.cc
new file mode 100644
index 0000000..eeb5168
--- /dev/null
+++ b/2023/puzzle-06-02.cc
@@ -0,0 +1,87 @@
+#include <cassert>
+#include <cmath>
+#include <cstdint>
+#include <cstdlib>
+#include <iostream>
+#include <iterator>
+#include <string>
+#include <string_view>
+#include <vector>
+
+using UInt = std::uint64_t;
+using NumVec = std::vector<UInt>;
+
+auto operator<<(std::ostream &os, NumVec const &vec) -> std::ostream & {
+  for (auto const num: vec) { os << ' ' << num; }
+  return os;
+}
+
+template<typename It>
+void read_nums(It it, std::string_view const nums) {
+  char const *pos{nums.data()};
+  while (pos - nums.data() < nums.size()) {
+    char *next_pos{nullptr};
+    *it++ = std::strtoull(pos, &next_pos, 10);
+    pos = next_pos;
+  }
+}
+
+[[nodiscard]] auto time_range(UInt time, UInt distance) -> UInt {
+  /* We hold down for `x` units of time and then travel (`t` - `x`) * `x` units of distance.
+   * So we want (t - x)x > d => tx - x² > d         (expand)
+   *                         => 0 > x² - tx + d     (shuffle)
+   *
+   * Lets solve the quadratic: x² - tx + d = 0
+   * with x = (-b ± √(b² - 4ac))/2a where a = 1, b = -t, c = d
+   *   => x = (t ± √(t² - 4d))/2
+   *   call two solutions x1 (negative root), x2 (positive root).
+   *
+   * Answer will the whole number of seconds between x1 and x2.  So:
+   *    floor(x2) - ceil(x1) + 1.
+   *
+   * However if x1 or x2 are the zero point we have to shift them too, so be careful of that.
+   */
+  double const t{static_cast<double>(time)};
+  double const d{static_cast<double>(distance)};
+  double const discriminant{std::sqrt(t * t - 4 * d)};
+  double const x1{(t - discriminant) / 2};
+  double const x2{(t + discriminant) / 2};
+  UInt x1i(std::ceil(x1));
+  UInt x2i(std::floor(x2));
+  if (x1i * x1i - time * x1i + distance <= 0) { ++x1i; }
+  if (x2i * x2i - time * x2i + distance <= 0) { --x2i; }
+  return x2i - x1i + 1;
+}
+
+auto main() -> int try {
+  std::string line;
+
+  NumVec times;
+  std::getline(std::cin, line);
+  assert(line.substr(0, 5) == "Time:");
+  std::erase(line, ' ');
+  read_nums(std::back_inserter(times), line.substr(5));
+
+  NumVec distances;
+  std::getline(std::cin, line);
+  assert(line.substr(0, 9) == "Distance:");
+  std::erase(line, ' ');
+  read_nums(std::back_inserter(distances), line.substr(9));
+
+  assert(distances.size() == times.size());
+
+  UInt score{1};
+  for (std::size_t i{0}; i != times.size(); ++i) {
+    auto const tr = time_range(times[i], distances[i]);
+    std::cout << times[i] << ' ' << distances[i] << " = " << tr << '\n';
+    score *= tr;
+  }
+
+  std::cout << "Score: " << score;
+
+  return EXIT_SUCCESS;
+}
+catch (...) {
+  std::cerr << "Uncaught exception.\n";
+  return EXIT_FAILURE;
+}