Solve for $v$.
$$2v^2-6v+29=(v-6{)}^2$$

Step 1 ：First, let's understand the problem. We are asked to solve for $v$ in the equation $2v^2-6v+29=(v-6{)}^2$.

Step 2 ：Expand the right side of the equation. $(v-6{)}^2$ becomes $v^2-12v+36$. So, the equation becomes: $2v^2-6v+29=v^2-12v+36$.

Step 3 ：Rearrange the equation to bring all terms to one side. This will give us a quadratic equation in standard form $a{x}^2+bx+c=0$. Subtract $v^2$, add $12v$, and subtract $36$ from both sides to get: $v^2+6v-7=0$.

Step 4 ：Factor the quadratic equation. The factors of -7 that add up to 6 are 7 and -1. So, the factored form of the equation is: $(v-1)(v+7)=0$.

Step 5 ：Set each factor equal to zero and solve for $v$. $v-1=0$ gives $v=1$ and $v+7=0$ gives $v=-7$. So, the solutions to the equation are $v=1$ and $v=-7$.

Step 6 ：Check the solutions. Substitute $v=1$ into the original equation: $2(1{)}^2-6(1)+29=(1-6{)}^2$ gives $25=25$ which is true. Substitute $v=-7$ into the original equation: $2(-7{)}^2-6(-7)+29=(-7-6{)}^2$ gives $169=169$ which is true. So, the solutions $v=1$ and $v=-7$ are correct.

Step 7 ：The final answer is $\boxed{{\displaystyle v=1}}$ and $\boxed{{\displaystyle v=-7}}$.