### Video Transcript

Find the solution set of the equation 𝑥 minus 23 all squared minus six 𝑥 is equal to zero, giving values to three decimal places.

We recall that any quadratic equation written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, can be solved using the quadratic formula. This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎.

Our first step in this question is to rewrite our equation so it is in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. In order to square 𝑥 minus 23, we multiply 𝑥 minus 23 by 𝑥 minus 23. We can distribute the parentheses or expand the brackets here by using the FOIL method. Multiplying the first terms gives us 𝑥 squared. 𝑥 multiplied by negative 23 is negative 23𝑥. Therefore, the product of the outer and inner terms are both negative 23𝑥. Finally, multiplying negative 23 by negative 23 gives us positive 529. Our equation becomes 𝑥 squared minus 23𝑥 minus 23𝑥 plus 529 minus six 𝑥 is equal to zero.

Next, we can collect the like terms. Our equation simplifies to 𝑥 squared minus 52𝑥 plus 529 is equal to zero. The values of 𝑎, 𝑏, and 𝑐 are one, negative 52, and 529. Substituting these into the quadratic formula gives us 𝑥 is equal to negative negative 52 plus or minus the square root of negative 52 squared minus four multiplied by one multiplied by 529 all divided by two multiplied by one. This simplifies to 𝑥 is equal to 52 plus or minus the square root of 588 all divided by two. This means that either 𝑥 is equal to 52 plus the square root of 588 all divided by two or 𝑥 is equal to 52 minus the square root of 588 all divided by two. Typing these calculations into our calculator and rounding to three decimal places, we get 38.124 and 13.876.

The solution set of the equation 𝑥 minus 23 all squared minus six 𝑥 equals zero contains two values: 38.124 and 13.876.