First manipulate data so that it will be in the for of Ax^2 + Bx + C = 0.
Then, you can do either of this:
1. Factor them out. Example:
6x^2 - x - 2 = 0
(3x -2)(2x + 1) = 0
3x - 2 = 0 2x + 1 = 0 Why??
x = 2/3 x = -1/2 Because when 1 of them becomes 0, then the whole equation becomes 0. Thus the zeroes/roots/solution set of the equation are/is {2/3, -1/2)
2. Use the quadratic formula
x =( -b +- √(b² - 4ac))/2a
Example: 6x^2 - x - 2 = 0
x = (-(-1) +- √((-1)² - 4(6)(-2)))/2(6)
x = (1 +- √49)/12
x = (1 +- 7)/12
x =( 1 + 7)/12 x = (1-7)/12
x = 8/12 x = -6/12
x = 2/3 x = -1/2
same results right? nut there ae times when you need to decide which one to use. there are some problems when it is better to use one over the other. however, not all equations are factorable, so knowing and udnerstanding the quadratic equation will really come in handy
