To solve this problem, let's consider the roots of the quadratic equation ax² + bx + c = 0 as tan@ and sec@.

1. Finding the quadratic equation:
Using the fact that the roots of a quadratic equation can be written as:
x = [-b ± √(b² - 4ac)] / (2a)

Substituting tan@ and sec@ as the roots, we have:
tan@ = [-b ± √(b² - 4ac)] / (2a)
sec@ = [-b ± √(b² - 4ac)] / (2a)

2. Simplifying the equations:
Recall that sec@ is the reciprocal of cos@, so we can rewrite sec@ as:
sec@ = 1 / cos@ = √(1 + tan²@)

Substituting this into the second equation, we have:
√(1 + tan²@) = [-b ± √(b² - 4ac)] / (2a)

Squaring both sides, we get:
1 + tan²@ = [(-b ± √(b² - 4ac)) / (2a)]²
1 + tan²@ = (b² - 2b√(b² - 4ac) + (b² - 4ac)) / (4a²)
1 + tan²@ = (2b² - 4ac - 2b√(b² - 4ac)) / (4a²)

3. Simplifying further:
Since tan²@ is equal to sec²@ - 1, we can rewrite the equation as:
sec²@ = (2b² - 4ac - 2b√(b² - 4ac)) / (4a²)
sec²@ = (b² - 2ac - b√(b² - 4ac)) / (2a²)

Recall that sec²@ is equal to 1 + tan²@, so we have:
1 + tan²@ = (b² - 2ac - b√(b² - 4ac)) / (2a²)

4. Rearranging the terms:
Multiplying both sides by a², we get:
a² + a²tan²@ = b² - 2ac - b√(b² - 4ac)

Rearranging the terms further, we have:
a²tan²@ + b√(b² - 4ac) = b² - 2ac - a²

5. The final relation:
Squaring both sides of the equation, we obtain:
a⁴tan⁴@ + 2a²b√(b² - 4ac)tan²@ + b²(b² - 4ac) = b⁴ - 4a²b² + 4a²c

Simplifying the equation, we get:
a⁴tan⁴@ + 2a²b√(b² - 4ac)tan²@ + b²(b² - 4ac) = b⁴ - 4a²(b² - c)

Since tan²@ is equal to sec²@ - 1, we can substitute this into the equation:
a