Does the equation \large x^2+12x+3=136 have the same solutions as the equation \large \left(x+6\right)^2=169 ?
First, we can simplify the second equation by taking the square root of both sides:
$$\begin{aligned} \ (x+6)^2&=169 \\ \ \sqrt{(x+6)^2}&=\sqrt{169} \\ \ x+6&=\pm13 \\ \ x&=-6\pm13 \end{aligned}$$
So the solutions to the second equation are $x=-19$ and $x=-7$.
Now, to see if these solutions also satisfy the first equation, we can substitute them in and see if the left-hand side equals $136$. For $x=-19$, we have:
$$x^2+12x+3=(-19)^2+12(-19)+3=361-228+3=136,$$
so $x=-19$ is a solution to the first equation. Similarly, for $x=-7$, we have:
$$x^2+12x+3=(-7)^2+12(-7)+3=49-84+3=-32,$$
so $x=-7$ is not a solution to the first equation.
Therefore, the two equations do not have the same solutions.