To solve $-3x^2 +2x +1=0$, I'd normally break the middle term and then factorise. But I was wondering if there was a way to skip the factorising step? The factors I'd use in place of the middle term would be $3$ and $-1$. If I were to flip the sign of each, I would get $-3$ and $1$. And then, were I to divide by the coefficient of $x^2$, I would get $1$ and $-1/3$, which are the correct answers. So, my question is, can all quadratics be solved in this way or was this just a fluke? If they can be, then why? If not, then is there any other quick and easy way of solving quadratics mentally? thanks for your help

What you've noticed is called the "Rational Root theorem". It says that if a polynomial equation has all integer coefficients, then any rational root of the equation has to have a numerator that divides the constant term, and a denominator that divides the leading coefficient. To use the theorem, you take all the divisors of the constant term, divide them by all the divisors of the leading coefficient, and then try the all the various results (and their opposites!) in the equation to see if any of them works. It is a fairly quick method, but alas it requires the polynomial to have integer coefficients, and there is no guarantee that it has any integer roots.

So no, you cannot rely on it as a means to solve quadratics. The quadratic equation is still your best friend here...

Notice, we have $$-3x^2+2x+1=0$$ Finding the roots of $-3x^2+2x+1=0$ by quadratic rule as follows

$$x=\frac{-2\pm\sqrt{(2)^2-4(-3)(1)}}{2(-3)}$$ $$x=\frac{-2\pm4}{-6}$$ $$x=\frac{-2+4}{-6}\ \ \vee \ \ x=\frac{-2-4}{-6}$$ $$x=1\ \ \vee \ \ x=-\frac{1}{3}$$ Then, $(x-1)$ & $\left(x+\frac{1}{3}\right)$ will be the factors, thus we can factorize the expression as follows $$-3x^2+2x+1=-3(x-1)\left(x+\frac{1}{3}\right)$$ Hence, we have $$-3(x-1)\left(x+\frac{1}{3}\right)=0$$ or $$(x-1)\left(3x+1\right)=0$$

Alternative Method Factorization can also be done by rearranging the terms as follows $$-3x^2+2x+1=0$$ $$-(3x^2-2x-1)=0$$ $$-(3x^2-3x+x-1)=0$$ $$-(\underbrace {3x^2-3x}+\underbrace{x-1})=0$$ $$-(3x(x-1)+(x-1))=0$$ $$(x-1)(3x+1)=0$$

Observe $a+b+c=-3+2+1=0 \Rightarrow 1$ is a root, thus $x-1$ is a factor of $p(x)=-3x^2+2x+1$, and since its a quadratic, you can guess the other factor: $-3x-1$.

Use the quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
So substitute the values and solve accordingly.
For derivation of it, $ax^2+bx+c=0$
$x^2+\frac{bx}{a}=-\frac ca$
$x^2+\frac{bx}a+\frac{b^2}{4a^2}=-\frac ca+\frac{b^2}{4a^2}$
$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

As Paul Sinclair brings up in his answer, by the Rational Root Theorem, it is only necessary to check $\pm1$ and $\pm\frac13$. You can use the theorem whenever all coefficients are integers, though of course if all coefficients are rational, you can always just multiply the entire equation by the lowest common denominator to make them integers.

However, it is possible the equation has no rational solutions or no real solutions. I did have a method to solve quadratics mentally, but I'll let others decide if it's easy.

You may already know the roots of any quadratic $ax^2+bx+c$ have a sum of $-\frac ba$ and a product of $\frac ca$. Let's say our two roots are $r+s$ and $r-s$. The sum of the roots is $2r$, so

$$r=-\frac b{2a}$$

Now the product of the roots is

$$r^2-s^2=\frac ca$$ $$s=\sqrt{r^2-\frac ca}$$

I find this a lot easier when $a=1$, $b$ is even, and $c$ is an integer, in which case you can avoid fractions. In your example though, we have $r=-\frac2{2\times-3}=\frac13$, so our answer is $\frac13$ plus or minus something. $s=\sqrt{(\frac 13)^2-\frac1{-3}}=\sqrt{\frac19+\frac13}=\sqrt\frac49=\frac23$. So the solution is $\frac13\pm\frac23$.

For a nicer example to test this method, try $x^2-8x+4=0$.