Hey all! im really stuck, i cant remember how to do this at all! and ive got several questions left!, help!

Solve the Following equations:

x² -2x -24 =0

And

(x²- x -30) (x - 7) = 0

2. I know the rule, i just dont know how to apply it? explain?

3. $x² -2x -24 =0$

For the 1st problem just factor or use the quad formula

$(x-6)(x+4) = 0$

Then solve for 0

$x-6 = 0$

$x+4 = 0$

x = -4 and 6

Do you understand what I did? Try doing the 2nd problem

4. Well i tried and failed lol. i see what you did in the first problem and i tried it on the other questions similar and it worked, but what happens when its factorised with x² in the bracket? like the second question?

5. ****! also what happens when theres two Squared numbers in both equationns, example $(x^2 - 5x -24) (x^2 +3x -28) =0$

6. Just keep factoring.

$(x² - x - 30) (x - 7) = 0$

$(x - 6)(x + 5)(x - 7) = 0$

Go ahead and try the other problem, I factored one out for you. See if you can do the rest

$(x^2 - 5x -24) (x^2 +3x -28) =0$

$(x - 8)(x + 3) (x^2 + 3x -28) = 0$