jmc

When we shift the graph of $y=2x^2-x+7$ four units to the right, we obtain the graph of $y=2(x-4{)}^2-(x-4)+7$, which simplifies to $y=2x^2-17x+43$. Therefore, $a+b+c=2-17+43=\boxed{28}$.

Another way to solve the problem is as follows: The graph of $y=a{x}^2+bx+c$ always passes through the point $(1,a+b+c)$. In other words, $a+b+c$ is the $y$-coordinate of the point on the parabola whose $x$-coordinate is 1. But this parabola is obtained by shifting the graph of $y=2x^2-x+7$ four units to the right, so $a+b+c$ is also the $y$-coordinate of the point on the original parabola whose $x$-coordinate is $1-4=-3$. This $y$-coordinate is equal to $2\cdot (-3{)}^2-(-3)+7=28$.