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
 A: This is called completing the square and there's a formula for how to do this.
Let $a=16$ and $b=56$. By our completing the square formula, we want:
$$c=\frac{b^2}{4a}=\frac{56^2}{4\cdot 16}=\frac{(8\cdot 7)^2}{2^2\cdot 4^2}=\frac{8^2\cdot 7^2}{2^2\cdot 4^2}=7^2=49$$
Thus, we want to change $-80$ to $+49$. We can do this by adding $+49-(-80)=129$ to both sides, getting us:
$$16x^2+56x-80+129=129$$
Simplify:
$$16x^2+56x+49=129$$
Now, we want to express the left side as a linear function squared. Since the $x^2$ term is $16x^2$, the $x$ term of the linear function must be $\sqrt{16x^2}=4x$. Since the constant term is $49$, the constant term of the linear function must be $\sqrt{49}=7$. Thus, we get:
$$(4x+7)^2=129$$
A: Hint:
Every equation of this form can be expressed in general form as $(lx+m)^2 +n$. Then compare coefficients of same powers.


$(lx+m)^2+n=l^2x^2 +2lmx+(m^2+n)=16x^2+56x-80$

*

*For $x^2$ term, $l^2=16 \Rightarrow l=\pm 4\tag{1}$

*For $x$ term, $2lm=56 \Rightarrow m=\pm 7\tag{2}$

*For constant term, $$m^2+n=-80 \Rightarrow n=-129\text{ [From $(2)$]}$$

Therefore, by choosing the positive solution $l=+4$ and $m=+7$, we get $lx+m=4x+7$, so:
$$16x^2+56x-80=(4x+7)^2-129$$
A: 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 $(4x- 7)^2- 129= 0$
A: 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
