<s>[INST] Convert $16x^2+56x-80=0$ to the form of $(\text{something})^2=D$ Sorry about this very basic question, I want to convert the 
equation $16x^2+56x-80=0$ to the form $(\text{something})^2=D$, I know that the answer is $(4x+7)^2-129$, but how can I convert this without usnig calculator
 [/INST] Here is an expansion which had helped me when I had come to know about completing the squares for the first time:
$ax^2+bx+c=0$
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + c = 0 $
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + (\frac{b}{2\sqrt{a}})^2 + c - (\frac{b}{2\sqrt{a}})^2 = 0$
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + (\frac{b}{2\sqrt{a}})^2  = (\frac{b}{2\sqrt{a}})^2 - c$
$(\sqrt{a}x+ \frac{b}{2\sqrt{a}})^2 = \frac{b^2}{4a} - c$
Clearly, $a \neq 0$ as this is a polynomial of second degree i.e. a quadratic equation. Hope it helps.
Cheers
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