Answer:
Part 1) The quadratic equation has zero real solutions
Part 2) The solutions are
and
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
The discriminant is equal to

If D=0 -----> the quadratic equation has only one real solution
If D>0 -----> the quadratic equation has two real solutions
If D<0 -----> the quadratic equation has two complex solutions
<em>Find the value of D</em>
-----> the quadratic equation has two complex solutions
<em>Find out the solutions</em>
substitute the values of a,b and c in the formula
Remember that

1427 = (F) (+ F + 60) (2F - 50) (+ 3F)
<em>Each pair of brackets represents one of the classes - F meaning Freshmen Class. It has been split into brackets for demonstrational purposes - nothing is being multiplied.</em>
F is an unknown number and each other class size is based off of that so we put it in algebraic terms in order to work it out.
1 - Freshmen Class
2 - Sophomore Class (Freshmen Class + 60 more students)
3 - Junior Class ( Twice the size of the Freshmen Class - 50 students)
4 - Senior Class (Three times the size of the Freshmen Class)
All this can be simplified to 7F + 10 = 1427
1427 - 10 = 1417
1417/7 = 202.428....
Is there a mistake in the question?
Following the question with 202 as the answer - the number 1424 is reached
If increased to 203 - 1431 is reached.
The answer shouldn't include half a person.