Select the correct answer. Complete the square for the following quadratic equation to determine its solutions in location of its extreme value. y=-x^2+4x+12 A. x=-2,6 extreme value at (2,12) B. x=2pm 2sqrt (7) extreme value at (2,16) C. x=2pm 2sqrt (7) extreme value at (2,12) D. x=-2,6 extreme value at (2,16) c ) A B

To complete the square for the quadratic equation $y=-x^{2}+4x+12$, we can rewrite it in the form $y=a(x-h)^{2}+k$, where $(h,k)$ represents the location of the extreme value.<br /><br />Step 1: Factor out the coefficient of $x^{2}$ from the first two terms:<br />$y=-(x^{2}-4x)+12$<br /><br />Step 2: Add and subtract the square of half the coefficient of $x$ inside the parentheses:<br />$y=-(x^{2}-4x+4-4)+12$<br />$y=-(x-2)^{2}+4+12$<br />$y=-(x-2)^{2}+16$<br /><br />So, the quadratic equation in completed square form is $y=-(x-2)^{2}+16$.<br /><br />The extreme value occurs at the vertex of the parabola, which is at $(h,k)=(2,16)$.<br /><br />Therefore, the correct answer is D. $x=-2,6$ extreme value at $(2,16)$.