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.

• 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. – Peter Jun 21 '19 at 10:53
• 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. – Arthur Jun 21 '19 at 10:53

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 $$\frac{-14}{17}$$.
$$(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2$$ 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.