General

r^3 + 3r^2 - 54r
r(r^2+3r-54)
r(r+9)(r-6)

c^2 - 49 <- Find the square root of 49
(c+7)(c-7)

x^2 - 4x - 12 = 0 <- You can chuck this into a scientific calculator and it'll give you an answer.
(x-6)(x+2)
x=6
x=-2

10x^2 = 9x
10x^2 - 9x = 0 <- Move the "9x" to the right-hand side and reverse the operation.
x(10x-9) = 0 -> x=0 (first value of x)
10x-9=0
10x=9
x=9/10 (second value of x)

I still don't understand how you're doing this, what is the formula for these equations?

[quote=Nivea]@kalvar: Oh me too. o.o
and we're learning the same material.....
do you happen to live in new jersey?[/quote]

Nope, weird coincidence though

[quote=Nivea]we're learning that right now,
what grade are you in if you dont mind me asking.[/quote]
Sophomore year of highschool in algebra 2

The second one:
(x-6)(x+2)=0
(x+2)=0, (x-6)=0
x=-2, x=6

10x^2-9x=0
x(10x-9) = 0
x=0, x= 9/10

c^2 - 49
(c + 7)(c - 7)

Whoah, can't believe I remember this stuff.

one sec, ill edit the answers in

R^3+3r^2-54r
1.r(r^2+3r-54)
2.r(r+9)(r-6)

c^2-49 Use difference of two squares

x^2 - 4x - 12 = 0
Unfactor (x-6)(x-2)=0 then solve each parenthesis seperately. (Each parenthesis equals 0)

10x^2 = 9x Subtract 9x from both sides, take a common factor, then solve the for the common factor and what's left after removing the common factor

Only have time for one:
r^3 + 3r^2 - 54r
factor out r
r(r^2 + 3r - 54)
Use quadratic formula on polynomial inside brackets or the other method you learned in class.