Solve for $x$: $(x+0.6)^2 = 1.4x^2$ I am very terrible at some aspects of algebra and I would like to ask how to solve this problem (It's actually only a small part of a larger physics problem). I've looked up the laws of exponents and from what I can tell I cannot easily seperate the $x$ from the $0.6$. At this point I am quite confused and would appreciate a step by step approach:

Solve for $x$:
$$(x+0.6)^2 = 1.4x^2$$

 A: $$(x+0.6)^2 = 1.4x^2$$
$$x^2+1.2x+0.36=1.4x^2$$
$$-0.4x^2+1.2x+0.36=0$$
Now use the ABC-formula to solve for x.
A: You can start by taking the square root of each side of the equation.
$$\sqrt{(x + 0.6)^2} = \sqrt{1.4x^2} =(x+0.6) = \sqrt{1.4}x$$
Then solve for $x$, and evaluate:
$$(\sqrt{1.4} - 1)x = 0.6 \implies x = \dfrac{0.6}{\sqrt{1.4} - 1} \approx 3.275$$
Note that this method gives you only the positive solution for $x$ (there is also a negative solution), but I suspect, given the solution you already have, is the only one applicable given the physic's context of the problem.

The alternative (and mathematically more sound) approach is to expand the term on the left hand side, and solve for the resulting quadratic equation:
$$
\begin{align}
(x+0.6)^2 &= 1.4x^2\\ \\
x^2+1.2x+0.36 &=1.4x^2\\ \\
-0.4x^2+1.2x+0.36&=0 
\end{align}
$$
Now, you can use the quadratic formula to obtain both the positive solution and the negative solution: $x \approx -0.275$ and $x\approx 3.275$.
A: $$(x+0,6)^{2}=1,4x^2$$
$$x^2+1,2x+0,36=1,4x^2$$
$$0,4x^2-1,2x-0,36=0$$
$$0,4(x^2-3x-9)=0$$
$$x^2-3x-9=0$$
then we get $x=\frac{3+3\sqrt{5}}{2}$ or $x=\frac{3-3\sqrt{5}}{2}$
i think it can be an answer without calculate the approximation value
A: $(x+0.6)^2 = 1.4x^2$ is not really a quadratic equation because $a^2=b^2$ iff $a=\pm b$, since $a^2-b^2=(a-b)(a+b)$.
Hence from $(x+0.6)^2 = 1.4x^2$ we get $x+0.6 = \pm \sqrt{1.4}\,x$ and so $x=0.6/(1\pm\sqrt{1.4})$.
