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Home » Frozen Tortillas: How to Perfectly Thaw and Use Them Every Week! - Anadea

Frozen Tortillas: How to Perfectly Thaw and Use Them Every Week! - Anadea

📅 May 17, 2026 👤 scraface

May 17, 2026

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📌 \frac{A + 2d}{A} = 1 + 2\frac{d}{A} = 1 + 2\frac{1}{x} = 1 + \frac{2x}{-21 \pm 3\sqrt{7}}.

📌 Take positive root for real chance (positive terms): $ x = \frac{-21 + 3\sqrt{7}}{14} $. Compute numerically: $ \sqrt{7} \approx 2.6458 $, $ 3\sqrt{7} \approx 7.937 $, so $ x \approx \frac{-21 + 7.937}{14} = \frac{-13.063}{14} \approx -0.932 $. Then $ \frac{1}{x} \approx -1.073 $, so $ 1 + 2(-1.073) = -1.146 $. Not meaningful.

📌 But from quadratic: $ 14x^2 + 42x + 27 = 0 $. Product of roots $ \frac{27}{14} $, sum $ -3 $.

📌 Let roots be $ r_1, r_2 $, then $ \frac{A + 2d}{A} = 1 + 2\frac{d}{A} = 1 + 2\frac{1}{x} $. Since product is $ x^2 = 28/9 $, $ x = \pm \sqrt{28}/3 $. Use symbolic. From $ 14x^2 + 42x + 27 = 0 $, divide by 14: $ x^2 + 3x + \frac{27}{14} = 0 $.

📌 Let $ y = \frac{1}{x} $, then from $ 14x^2 + 42x + 27 = 0 $, divide by $ x^2 $:

📌 $ 14 + 42y + 27y^2 = 0 \Rightarrow 27y^2 + 42y + 14 = 0 $.

📌 y = \frac{-42 \pm \sqrt{1764 - 1512}}{54} = \frac{-42 \pm \sqrt{252}}{54} = \frac{-42 \pm 6\sqrt{7}}{54} = \frac{-7 \pm \sqrt{7}}{9}.

📌 We want positive ratio. Try $ y = \frac{-7 + \sqrt{7}}{9} \approx \frac{-7 + 2.6458}{9} = \frac{-4.3542}{9} \approx -0.484 $, negative. Other root: $ \frac{-7 - 2.6458}{9} < 0 $. Both negative — meaning $ A/d < 0 $, so first and last have opposite signs — impossible for $ A^2 + (A+3d)^2 = (sum)^2 $ unless not ordered.

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