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Solving an Absolute Value Radical Equation

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The steps below outline a method for solving the following equation for tt.
25∗t2+10∗t+1−5∗t+2t2+0.4∗t+0.04=∑n=11n\sqrt{25*t^2+10*t+1}-\frac{5*t+2}{\sqrt{t^2+0.4*t+0.04}}=\sum_{n=1}^{1} n

Step Description
  1. | The expressions under the radicals are perfect squares: 25∗t2+10∗t+1=∣5∗t+1∣\sqrt{25*t^2+10*t+1}=|5*t+1| and t2+0.4∗t+0.04=∣t+0.2∣\sqrt{t^2+0.4*t+0.04}=|t+0.2|.
  2. | The equation can be rewritten using absolute values: ∣5∗t+1∣−5∗t+2∣t+0.2∣=1|5*t+1|-\frac{5*t+2}{|t+0.2|}=1.
  3. | Assuming 5∗t+1>05*t+1>0 and t+0.2>0t+0.2>0, the equation simplifies to the quadratic 5∗t2−4∗t−2=05*t^2-4*t-2=0.

Which of the following is a solution for tt that satisfies the assumptions made in Step 3?

A

t=−2−145t=\frac{-2-\sqrt{14}}{5}

B

t=2+145t=\frac{2+\sqrt{14}}{5}

C

t=−2+145t=\frac{-2+\sqrt{14}}{5}

D

t=2−145t=\frac{2-\sqrt{14}}{5}

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