Question Show Examples Solve for the roots in simplest form by completing the square: \[ 3 x^{2}-60 x+741=0 \]

Step 1 ：Given the quadratic equation \(3x^{2}-60x+741=0\)

Step 2 ：First, we make the coefficient of \(x^{2}\) equal to 1 by dividing the entire equation by the coefficient of \(x^{2}\), which is 3. This gives us \(x^{2}-20x+247=0\)

Step 3 ：Next, we move the constant term to the right side of the equation, resulting in \(x^{2}-20x=-247\)

Step 4 ：We then take half of the coefficient of x, which is -20, square it, and add it to both sides of the equation. This gives us \(x^{2}-20x+100=347\)

Step 5 ：The left side of the equation is now a perfect square. We can write it as a square of a binomial, resulting in \((x-10)^{2}=347\)

Step 6 ：We take the square root of both sides of the equation, giving us \(x-10=\pm\sqrt{347}\)

Step 7 ：Solving for x, we get \(x=10\pm\sqrt{347}\)

Step 8 ：However, since \(\sqrt{347}\) is not a perfect square, we simplify it to \(x=10\pm7\sqrt{3}\)

Step 9 ：Finally, we find that the roots of the equation are complex numbers. The real part of the roots is 10 and the imaginary part is ±7√3

Step 10 ：Final Answer: The roots of the equation are \(\boxed{10 - 7\sqrt{3}i}\) and \(\boxed{10 + 7\sqrt{3}i}\)