For this problem, we are asked to provide the intermediate step we obtain while completing the square for the following expression:
[tex]-6x^2-235=-48x+11[/tex]Completing the square means to transform the function into a expression such as:
[tex](a+b)^2=a^2+2\cdot a\cdot b+b^2[/tex]For this we will first change all the terms to the left side, as shown below:
[tex]\begin{gathered} 6x^2-48x+235+11=0 \\ 6x^2-48x+246=0 \end{gathered}[/tex]Since all three terms are divisible by 6, we need to use factorization to isolate the 6 from the equation:
[tex]6\cdot(x^2-8x+41)=0[/tex]Now, we need to rewrite the expression inside the parenthesis, such as we will obtain a form that is roughly equal to the perfect square we're looking for.
[tex]\begin{gathered} 6\cdot(x^2-2\cdot4\cdot x+41)=0 \\ 6\cdot(x^2-2\cdot4\cdot x+16+25)=0 \end{gathered}[/tex]Now, we need to remove the "25" from the parenthesis, for that we need to multiply the 6 by 25.
[tex]\begin{gathered} 6\cdot(x^2-2\cdot4\cdot x+16)+6\cdot25=0 \\ 6\cdot(x^2-2\cdot4\cdot x+16)+150=0 \\ 6\cdot(x^2-2\cdot4\cdot x+16)=-150 \end{gathered}[/tex]Now we can transform the parenthesis to the sum of two squares, where the term "a" is equal to x, and the "b" is equal to 4.
[tex]6\cdot(x^{}-4)^2=-150[/tex]