(x-24)^2=18. I was wondering how you solve this problem?

Simply expand the brackets, transpose the 18 to the LHS and apply a suitable method of solution of quadratic equations. That is how we solve this (and all such) equation. Why don't you make a try?

There is more than one way to do it... but here is one way

(x-24)^2=18
(x-24)= +/-root(18) = +/-root(9x2) = +/- 3 root(2)
x = 24 +/- 3 root(2)

(x - 24)^2 = 18
x - 24 = ±√18
x - 24 = ±√(3^2 * 2)
x = 24 ± 3√2

x^2-48x+576=18

x^2-48x+558=0


a=1 b=-48 c=558

-(-48)+/- sq rt -48^2-4(1)(558)/2
48+/- sq rt 2304- 2232/ 2
48+/- sq rt 72/ 2
48+/- sq rt 36 sq rt 2 / 2
48+/- 6 sq rt 2 / 2

x= 24+/- 3 sq rt 2

x=24 + 3 sq rt 2 x= 24- 3 sq rt 2

Substitute (x - 24) with u to produce u^2 = 18.
Try factoring 18 such that one of its factors is a square. In this case: 2 x 9.
With the 2x9 clue, you can square root further 9, to produce 3. This is your whole number.
Put 2 inside a radical symbol. This produces either 3sqrt(2) or -3sqrt(2).
With the two half-answers, just add 24 to both half-answers to get 2 answers.
The answers are: {24 + 3sqrt(2),24 - 3sqrt(2)}

x-24 = +/- 3rt2
So x = 24+3rt2 or 24-3rt2.

(x-24)(x-24)=18
x^2-48x+576=18
x^2-48x+558=0

That doesnt factor so do the quadratic formula.
and you end up with 24 +- 3Root2 which is about 28.23 and 19.77

(x-24)^2=18
=>x^2-48x+576=18
=>x^2-48x+576-18=0
=>x^2-48x+558=0

x= 28.24 &19.75

square root each side then get
x-24 = spare root of 18
subtract 24
x = square root of 18 - 24

(x-24)^2=18
<=>
x-24 = +- sqrt(18)
<=>
x = 24 +- sqrt(18)

So you got that 2 solutions.