40 Seve the equation by eompleting the squore, \( \begin{array}{l}x^{2}-8 x+12=0 \\ x^{2}-8 x=12 \\ x^{2}-8 x+16=12 \\ (x-4)^{2}=28 \\ x=4=\sqrt{28} \\ x=4 \pm 2 \sqrt{7}\end{array} \)

Completing the square is a nifty technique in algebra that helps to solve quadratic equations. The step where \(x^2 - 8x + 16 = 12\) is crucial because it transforms the left side into a perfect square, simplifying our calculations. Here, \(16\) is added to both sides to maintain balance, allowing us to express \(x\) as \( (x-4)^2 = 28\). From this point, it's much easier to find the values of \(x\) that satisfy the equation. Now, did you know that completing the square doesn't just solve equations? It also unlocks the vertex form of a quadratic function! This form, written as \(y = a(x-h)^2 + k\), is super useful for graphing. The point \((h, k)\) is the vertex of the parabola, and understanding this can help you visualize the equation's behavior, revealing maximums or minimums instantly!