quadratic equation solving mistake I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
\begin{align}
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\\
8 + 16x^2 + x &= 4x - x^2 + 22\\
16x^2 + x &= 4x - x^2 + 14\\
16x^2 &= 3x - x^2 + 14\\
17x^2 &= 3x + 14
\end{align}
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.

You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $\frac{-14}{17}$.
