Given the algebraic equation
[tex]v^2-12v+28=-7[/tex]We are asked to solve the question using the completing the square method. This can be done with the following steps
Explanation
Step 1: Take all the constant to the right hand side
[tex]\begin{gathered} v^2-12v+28=-7 \\ v^2-12v=-7-28 \\ v^2-12v=-35 \end{gathered}[/tex]Step 2: Add to both sides the square of half the coefficient of x
[tex]\begin{gathered} v^2-12v+(-\frac{12}{2})^2=-35+(-\frac{12}{2})^2 \\ \end{gathered}[/tex]Step 3: Factorize the left-hand sign and simplify the right-hand side.
[tex]\begin{gathered} v^2-12v+(-\frac{12}{2})^2=-35+(-\frac{12}{2})^2 \\ v^2-12v+6^2=-35+6^2 \\ (v-6)^2=1 \end{gathered}[/tex]Step 4: Find the square root of both sides
[tex]\begin{gathered} \sqrt[]{(v-6)^2}=1 \\ v-6=\pm\sqrt[]{1} \\ v=6\pm\sqrt[]{1} \end{gathered}[/tex]The solution to the equation is
Answer:
[tex]v=7,\: v=5[/tex]