Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$? 
 A: $$3\left(2x-1\right)^{2}=147\iff\left(2x-1\right)^{2}-49=0\iff$$$$\left[\left(2x-1\right)-7\right]\left[\left(2x-1\right)+7\right]=0\iff\left(2x-8\right)\left(2x+6\right)=0$$
A product $p\times q$ equals $0$ if and only if $p=0$ or $q=0$.
Applying that here gives $2x-8=0$ or $2x+6=0$.
A: If $3(2x-1)^2=147=3\times7^2$, then you have an equation of the form $A^2=B^2$ with $A=2x-1$ and $B=7$, hence $A=\pm B$:
You can write that $A^2=B^2$ is equivalent to $A^2-B^2=0$, but $A^2-B^2=(A-B)\cdot(A+B)$, hence it's also equivalent to $(A-B)\cdot(A+B)=0$. But a product is zero if and only if at least one of the factor is zero, hence $A-B=0$ or $A+B=0$, that is, $A=B$ or $A=-B$. 
Now, replace $A$ with $2x-1$, and $B$ with $7$, and you get:
$$2x-1=\pm 7$$
Which is an abuse of notation for $$(2x-1=+7)\ \ \ \mathrm{ or }\ \ \ (2x-1=-7)$$
So you have the two solutions $x=4$ and $x=-3$:
$$2x-1=+7\implies 2x=8\implies x=4$$
$$2x-1=-7\implies 2x=-6\implies x=-3$$
A: $$ 3(2x-1)^2=147 $$
$$ (2x-1)^2=\frac{147}{3} $$
$$ (2x-1)^2=49 $$
$$ (2x-1)^2=(\pm 7)^2 $$
$$ \sqrt{(2x-1)^2}=\sqrt{(\pm 7)^2} $$
$$ 2x-1=\pm 7$$
$$ 2x=1\pm 7$$
$$ x=\frac{1\pm 7}{2}$$
Therefore 
$$ x_1=\frac{1+ 7}{2}=\frac82=4$$
$$ x_2=\frac{1- 7}{2}=-\frac62=-3$$
