### Video Transcript

Find the solution set of the
equation π₯ minus five squared minus π₯ minus five minus three equals zero, giving
values to one decimal place.

Now, what weβre not actually going
to do is distribute our parentheses and simplify. Instead, we need to spot that we
already have a quadratic equation of some from. If you need convincing of this,
letβs perform a substitution. Weβre going to let π¦ be equal to
π₯ minus five. Then, π₯ minus five all squared is
π¦ squared. Negative π₯ minus five becomes
negative π¦. And then, we subtract three. So our equation becomes π¦ squared
minus π¦ minus three equals zero.

Our job is to solve for π¦. Once we have values for π¦, we can
use our earlier substitution to find the suitable values for π₯. So how do we solve this quadratic
equation? Well, we canβt factor the
expression π¦ squared minus π¦ minus three. So we have two options. We can either complete the square
or we can use a quadratic formula. For a quadratic equation of the
form ππ¦ squared plus ππ¦ plus π equals zero, the solutions are given by π¦
equals negative π plus or minus the square root of π squared minus four ππ over
two π. π is the coefficient of π¦
squared, so itβs one. π is the coefficient of π¦, so
thatβs actually negative one. And π is the constant; itβs
negative three.

We substitute everything we have
into our quadratic formula. And we get π¦ is equal to negative
negative one plus or minus the square root of negative one squared minus four times
one times negative three all over two times one. This simplifies to π¦ equals one
plus or minus the square root of 13 over two. So we have two solutions for
π¦. We have one plus the square root of
13 over two and one minus the square root of 13 over two. But what are our solutions for
π₯?

Well, we go back to our earlier
substitution, we said π¦ was equal to π₯ minus five. By adding five to both sides, we
find that π₯ is equal to π¦ plus five. And so for the solution π¦ equals
one plus the square root of 13 over two, π₯ is that value plus five. And the other solution for π₯ is
one minus the square root of 13 over two plus five. That gives us 7.3027 and so on or
3.6972 and so on. We round each number to one decimal
place. And then, we use these little curly
brackets to represent the solution set. And our solution set is 7.3 and
3.7.