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. $\displaystyle x² -2x -24 =0$

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

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

Then solve for 0

$\displaystyle x-6 = 0$

$\displaystyle x+4 = 0$

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

6. Just keep factoring.

$\displaystyle (x² - x - 30) (x - 7) = 0$

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

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

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

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