Now, one complication.  
In this example, I deliberately made up the example so that it could be factored easily.
That will not always be the case.
So, if you can’t factor a quadratic equation, you can always use the quadratic formula.  
Now, let me just give the quadratic formula to you.
And let’s switch from P to x because the quadratic formula is given in terms of x instead of P.
Let’s assume you have an equation of this form: a x squared plus bx plus c = 0.
Where the variable here, your variable is x.  
And a, b and c are parameters.
And there’s no restriction really on a, b, and c other than, in order for this to be a quadratic equation, we don’t want a to be zero. 
If a is zero, then notice what kind of equation is it?  
If a were zero, what kind of equation would this be?  
It would be a linear equation.
.  So, the only restriction really is a is not zero.  
In other words, x squared appears in your expression.  
So, let’s just assume we have that restriction:  a is not equal to zero.  
It could be positive or negative.
Whenever you have an equation of this form, you can work it out algebraically.
And the derivation is given in the book.  
There are two roots, two solution values for x.  
They are x one star, x two star equals minus b plus or minus the square root of b squared minus 4 ac all over 2a.
Now, in our problem right here, we get P instead of x.  
.  But just making that change, what is our value of a?
1.
So, we have a equals 1 in the problem we just did. 
What is our value of b?  
10.
And what is our value of c?  
Negative 200.
So, if you had a problem like this and you couldn’t see how to factor it, then you could just the use the quadratic formula.  
This formula right here.
And notice, if I just substitute into the quadratic formula,
instead of x one star, x two star, it’s P one star, P two star,
equals minus b, so it’s minus 10.  
Plus or minus the square root,
b squared is going to be 100.  
10 squared is 100.  
Minus 4. 
A is equal to 1.  
C is equal to minus 200,
all divided by 2.
So, if we just continue this.  Notice what we get.  
We get minus 10
plus or minus
And the plus or minus gives us the two different roots.
.  One for plus, the other for negative.  
So, this is 100
this is -4 times a 
a -200 is 800.  
So, we have the square root, one square root, if we added, is 900 over 2.
The square root of 900 is what?
30.  
So, if we add this, -10 plus 30, that would be 20 over 2.
That would be our 10.  Okay.
If we subtracted it, 
-10 minus 30 would be -40, over 2 would be -20. 
So you see, if we disregard this one, this negative value for price, we get the same answer. 
P equals 10.  Okay.
So, you can solve these nonlinear models using the same process of elimination of variables.  
It’s just a little more tedious to do.
