How do you solve y equals 3 x plus 5 x minus 2 after the 3 there are parentheses between x plus 5 and x minus 2?

y = 3(x+5)(x-2)

y = 3 (x^2 +2x + 5x + 10)

y = 3 (x^2 +7x + 10)

y = 3(x^2) +21x + 30
What are you trying to solve exactly?

Are you looking now for X based on Y ?

The problem is that there are 2 unknown variables and only 1 equation, so you can only solve for one of the variables based on the other.

Since you already have what Y is equal to, I suppose I'll just solve for X based on Y using the quadratic formula.

Now the quadratic formula only works when we set it to equal 0, so the first step is to move the Y to the other side.


3(x^2) +21x + (30-y) = 0


x =[ -b +/- sqrt( b^2 - 4ac) ] / 2a

where

a = 3
b = 21
c = (30-y)

So, plugging in out values into the equation, we get:

x = [ -21 +/- sqrt( 21^2 - 4*3*(30-y)) ] / (2*3)

?x = [ -21 +/- sqrt(441 - 360 +12y) ] / 6

Now you just plug in whatever value you want into Y and you can compute X.

Y is the independent variable in this case.