<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] You are "completing the square".  The firs thing I would do is factor "16" out of the first two terms: $16(x^2+ \frac{7}{2}x)- 80= 0$. The reason I did that I that I know that a "perfect square" is of the form $(x+ a)^2= x^2+ 2ax+ a^2$. Compare that to $x^2+ \frac{7}{2}x$.  The first two terms  will be the same  if $2a= \frac{7}{2}$ or $a= \frac{7}{4}$.  In that case, $a^2= \left(\frac{7}{4}\right)^2= \frac{49}{16}$.  We can get a "perfect square" by adding $\frac{49}{16}$.  Of course, in order not to change the value, we have to subtract that also:
$16(x^2+ \frac{7}{2}x+ \frac{49}{16}- \frac{49}{16})- 80= 16(x^2+ \frac{7}{2}x+ \frac{49}{16})- 49- 80= 16(x- \frac{7}{4})^2- 129$.
Putting the "16" backinside the square will give $(4(x- \frac{7}{4}))^2= (4x- 7)^2$ so we have $(4x- 7)^2- 129= 0$
</s>