Form and solve the quadratic equation: “Five times a positive integer equals 3 less than twice the square of that integer.” Find the integer.

Introduction / Context:
This is a classic translate-to-equation problem. Converting the sentence into an algebraic equation and then solving the resulting quadratic will yield the required positive integer. Such questions assess equation formation and careful arithmetic.

Given Data / Assumptions:

• The unknown is a positive integer n.
• Statement: 5*n equals 2*n^2 − 3.
• We seek the positive integer solution.

Concept / Approach:
Translate the sentence into an equation, collect like terms, and solve the quadratic using factoring or the quadratic formula. Check which root is positive and integral, as required by the problem statement.

Step-by-Step Solution:
Write the equation: 5n = 2n^2 − 3.Rearrange: 2n^2 − 5n − 3 = 0.Compute discriminant: D = (−5)^2 − 4*2*(−3) = 25 + 24 = 49.Roots: n = [5 ± 7] / (2*2) = (5 ± 7)/4 → n = 12/4 = 3 or n = −2/4 = −0.5.Only the positive integer solution is allowed → n = 3.

Verification / Alternative check:
Test n = 3 in the original sentence: LHS = 5*3 = 15; RHS = 2*(3^2) − 3 = 18 − 3 = 15. Equality holds.

Why Other Options Are Wrong:
13, 23, 33 do not satisfy the equation; “None of these” is incorrect because n = 3 works exactly.

Common Pitfalls:
Sign mistakes when moving terms; arithmetic errors on the discriminant; forgetting to reject non-integer or negative roots.

Final Answer:
3