Me i'm a desperate sobbing mess cause of this class. solve by factoring 1) 2z^2-36=0 2) c^2=5c 3) p^2-4p=21 4) 2w^2-11w= -12 write the equation in standard form. then solve. 5) 4q^2+3q^2-4q=18

Don't sob! I gotchu!

2. c^2=5c aka c^2-5c+0=0 right?
now we need a number on the left which will multiply to give c^2, we will get that by c*c good?
now to the right side, we need numbers that multiply to give 0 and sum to -5, we'll get that by -5*0=0 and -5+0=-5 good?
now you get the factor (c-5)!

now this way, while factoring an equation that has a coefficient of more than 1, we multiply that coefficient to the constant (in the first question, 2 will be multiplied to 36 giving 72) now we need numbers that multiply to give 72 and sum to 0, since there isn't a number like that, we will have to use the quadratic formula. get it?

(also if the alphabets get too confusing, you can always use x, it doesn't matter!)