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

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:
$$(4x+7)^2=129$$
