In part A, we simply replace the variable A with 326
This is because A represents the number of vehicles in thousands.
So we go from this [tex]A = 2x^2+22x+326[/tex] to this [tex]326 = 2x^2+22x+326[/tex]
Answer: [tex]326 = 2x^2+22x+326[/tex]
==========================================
For part B, we'll subtract 326 from both sides
[tex]326 = 2x^2+22x+326\\\\326-326 = 2x^2+22x+326-326\\\\0 = 2x^2+22x\\\\2x^2+22x = 0[/tex]
Then we can factor out the GCF 2x and solve for x.
[tex]2x^2+22x = 0\\\\2x(x+11) = 0\\\\2x = 0 \text{ or } x+11 = 0\\\\x = 0 \text{ or } x = -11[/tex]
What does this tell us? It says that for years x = 0 and x = -11, the amount of vehicles purchased was A = 326 thousand.
We ignore x = -11 because that would mean we rewind 11 years into the past, but the instructions say "x years after 1980". In other words, 1980 is the farthest in time we can go back. That corresponds to the year x = 0.
In short, the only practical solution here is x = 0. It says 326 thousand cars were purchased in 1980.
Answer: 1980
