Question

Determine the solution set of (x-4)^2=12.A: {4+2 square root of 3, 4-2 square root of 3}B: {4-2 square root of 3, 4-2 square root of 3}C: {2 square root of 3 + 4, 2 square root of 3 -4}

Answer 1

$\begin{gathered} (x-4)^2=12 \\ x^2-8x+16=12 \\ x^2-8x+16-12=0 \\ x^2-8x+4=0 \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=1 \end{gathered}$