State 2 values of $x$ for which the value of $3x^2+4x-14$ is a perfect square. State 2 values of $x$ for which the value of $3x^2+4x-14$ is a perfect square.


I can't seem to factor it and I'm really lost. Does using the quadratic equation for it work? Help I need it for grade 10 quadratics test.
$$(1)\quad3x^2+4x-14=a^2$$ $$\implies 3x^2+4x-(14+a^2)=0\implies x=\dfrac{-4\pm\sqrt{16-12(-14-a^2)}}{6}$$

This does not need to be an integer. This says that for $(1)$ to be true, just choose an $a$ here and it will give you an $x$ value that satisfies. For example, $a=1$: $$x=\dfrac{-4\pm\sqrt{16-12(-15)}}{6}=\dfrac{-4\pm\sqrt{196}}{6}=\frac{10}{6},-3$$
Plugging either one of these values into the original equation will give you $a=1$.
3x^2+4x-14=0.
Let q and w be 2 roots of the equation.
q+w=4,qw=-42
(q-w)^2=(q+w)^2-4qw
So q-w=√184=2√46
On solving q+w=4and q-w=2√46 
We get w=q={4+2√46}÷\2
Now put the value of w andq and solve it.on verifying q+w is not equal with 4 . slight deviations will come as we have taken imaginary roots.