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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.

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    $\begingroup$ Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-\frac{14}{17}$ to avoid confusion. $\endgroup$ – Peter Jun 21 '19 at 10:53
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    $\begingroup$ When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions. $\endgroup$ – Arthur Jun 21 '19 at 10:53
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Your working is all fine. Only the last step was remaining to be written down after factoring.

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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 $\frac{-14}{17}$.

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$$ (1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2 $$ seems true ($25$ in both members), and with a little more effort $$ \left(1 - 4\frac{\overline{14}}{17}\right)^2 + 9\frac{\overline{14}}{17} + 7 = 2\left(\frac{\overline{14}}{17}+3\right)\left(1-\frac{\overline{14}}{17}\right) + \left(\frac{\overline{14}}{17}+4\right)^2 $$ is

$$(17+56)^2-9\cdot17\cdot14+7\cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.

As a quadratic equation has at most two roots, your work is right.

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