Given that the line
$3x + y -1 = 0$
is tangent to the parabola with equation
$y = -2x^2 +px + (1-p)$
EDIT: P > 3 EDIT 2: Fixed equation of tanget
I have to find $p$. I have tried making them equal to find a common point but that didn't turn out anything useful, any help is much appreciated.
The line and the parabola have to meet at a single point, hence the discriminant of the second-degree polynomial $$ -2x^2+px+(1-p)-(3x-1) = -2x^2 + (p-3) x + (2-p) $$ has to be zero. That implies $(p-3)^2+8(2-p)=0$ and $p=\color{red}{7\pm2\sqrt{6}}$.
Hint: if you substitute the linear equation ($y=3x-1$) into the parabola you get $$2x^2+(3-p)x+(p-2)=0.$$So now you have a quadratic equation that can only have one zero, since the line is tangent to the parabola. What does this tell you about the discriminant?
This is after you've corrected the equation of your tangent (Edit 2):
Well we can substitute the equation of the parabola into the equation of the tangent to give: $$3x-2x^2+px+(1-p)-1=0$$ $$-2x^2+(p+3)x-p=0$$ Note that the discriminant $\Delta$ for a repeated root must equal zero. Therefore, you must set the discrimant equal to zero, and solve for $p$. $$\Delta=(p+3)^2-4\cdot2\cdot p=0$$ The solutions of $p$ are complex, indicating there does not exist a real value of $p$ such that the line is tangent to the parabola.
Try varying $p$ using the slider with Desmos Graphing Calculator, and you will see what I mean.