Solve the equation completing the square t^2-10t+24=0?

To complete the square, you need to add and subtract 1 to
the left side of the equation:

t^2 - 10t + 24 + 1 - 1 =
0

t^2 - 10t + 25 - 1 = 0

The
first three terms are the terms of a perfect square:

(t -
5)^2 - 1 = 0

The difference of squares returns the
product:

x^2 - y^2 =
(x-y)(x+y)

Let x = t - 5 and y =
1

(t - 5)^2 - 1 = (t - 5 - 1)(t - 5 +
1)

(t - 5)^2 - 1 =
(t-6)(t-4)

But (t - 5)^2 - 1 = 0 => (t-6)(t-4) =
0

We'll set each factor as
zero:

t - 6 = 0 => t =
6

t - 4 = 0 => t =
4

The solutions of the equation are: {4 ;
6}.