Let $f\left(x\right)=3x-2x^2+1$

Differentiating both sides w.r.t. $x$, we get

$f^{\text{'}}\left(x\right)=3-4x$

For extreme value,

$f^{\text{'}}\left(x\right)=0$

$\Rightarrow 3-4x=0$

$\Rightarrow x=\frac{3}{4}$

So,

$f\left(\frac{3}{4}\right)=k$

$\Rightarrow 3\times \frac{3}{4}-2\times \frac{9}{16}+1=k$

$\Rightarrow 36-18+16=k$

$\Rightarrow k=34$

So, $34x^2+2x+1>0 \cdots \left(\mathrm{i}\right)$

But discriminant of $34x^2+2x+1$ is $D=4-136=-132<0$

And, coefficient of $x^2$ is greater than zero, so $\left(\mathrm{i}\right)$ is true for all $x\in R$, hence required solution is $\left(-\infty ,\infty \right)$.