If the polynomial $p\left(x\right)=x^2+3x-1$ is such that $p\left(x\right)=p(1-x),$ then the value of $\text{'}x\text{'}$ is 

$p\left(x\right)=p(1-x)$

$x^2+3x-1={(1-x)}^2+3(1-x)-1$

$x^2+3x-1=(1+x^2-2x)+(3-3x)-1$

$x^2+3x-1=1+x^2-2x+3-3x-1$

$3x-1=-5x+3$

$3x+5x=3+1$

$8x=4$

$x=\frac{4}{8}$

$x=\frac{1}{2}$