Algebra 2 help！！！！！！！Please！！！！(roots)

Using quadratic formula, solution to ax² + bx + c = 0 is
x = (−b ± √(b²−4ac)) / (2a)

Value under square root is called the discriminant (D)
D = b²−4ac

x = (−b ± √D) / (2a)


When D < 0, then √D is not real.
So x = (−b ± √D) / (2a) is not real.

Therefore, to get real roots, we need D ≥ 0


When D = 0, then we get: x = (−b ± √0) / (2a) = −b/(2a)
So we get two roots with the same value (i.e. a double root)

Therefore, to get real and unequal roots, we need D ≥ 0 and D ≠ 0, i.e. D > 0


When D is not a perfect square, then √D is not rational,
So x = (−b ± √D) / (2a) is not rational

Therefore, to get real, rational, and unequal roots, we need D = perfect square

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x² + x + 1 = 0 -------> D = 1² − 4 = −3 < 0 -------> non-real roots
x² − 4x + 4 = 0 -----> D = 4² − 4(4) = 0 ----------> double non-unique roots
x² − 4 = 0 -----------> D = 0 − 4(1)(−4) = 16 ----> real, rational, and unequal roots
x² − 2 = 0 -----------> D = 0 − 4(1)(−2) = 8 -----> real, irrational, and unequal roots

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ax² + 6x − 9 = 0

Imaginary root are generated if D < 0
6² − 4(a)(−9) < 0
36 + 36a < 0
36a < −36
a < −1

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Find x so that p(x) > 0

I'll leave it to you to complete and won't bother doing this for you because:
1. profit function is usually an inverted parabola (not the upright parabola you have)
2. p(x) = x^2 = 120x - 200 is not a function ("=" sign in front of 120x ????)
3. I'm not sure what those answer choices mean (why is there an x at the beginning of each answer choice, and why do you have 120 < x < 100 ???)