Solve the following equations

7f+f^2=60
6x^2-x-15=0
6(y-3)^2-(y-3)-15=0

1 answer

always rearrange things into descending powers. Set equal to zero because after you factor it, you know that the product can be zero only if one of the factors is zero.

f^2+7f-60 = 0
(f+12)(f-5) = 0
f = -12 or 5

6x^2-x-15 = 0
(2x+3)(3x-5) = 0
x = -3/2 or 5/3

6(y-3)^2 - (y-3) - 15 = 0
(2(y-3)+3)(3(y-3)-5) = 0

one way: y-3 = -3/2 or 5/3
That means y = 3/2 or 14/3

other way:
(2y-6+3)(3y-9-14) = 0
(2y-3)(3y-14) = 0
y = 3/2 or 14/3