Question 18 Find all real solutions of the equation by completing the square \( x^{2}-6 x+3=43 \). \( x_{1}=\square \) and \( x_{2}=\square \) with \( x_{1} \) Next Question

To solve the equation \( x^{2}-6x+3=43 \), we first bring all terms to one side to set the equation to zero: \( x^{2}-6x+3-43=0 \) or equivalently, \( x^{2}-6x-40=0 \). Next, we complete the square: 1. Take half of -6, which is -3, and square it to get 9. 2. Rewrite the equation as \( (x-3)^{2} - 9 - 40 = 0 \), simplifying to \( (x-3)^{2} - 49 = 0 \). 3. This gives \( (x-3)^{2} = 49 \). Taking the square root of both sides results in two equations: 1. \( x-3 = 7 \) leading to \( x = 10 \), 2. \( x-3 = -7 \) leading to \( x = -4 \). Thus, the solutions are \( x_{1} = -4 \) and \( x_{2} = 10 \).