Hello, Raj!
1. Solve algebraically
$\displaystyle A)\;\;2x^2+5x63\:>\:0$
The graph of $\displaystyle y \:=\:2x^2+5x63$ is an upopening parabola.
. . We want to know when the graph is above the xaxis.
The function is positive to the left and the right of its xintercepts.
xintercepts: .$\displaystyle 2x^2+5x63 \:=\:0\quad\Rightarrow\quad (x + 7)(2x  9)\:=\:0\quad\Rightarrow\quad x \:=\:7,\:\frac{9}{2}$ Code:

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o+o
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Answer: .$\displaystyle \boxed{x < 7\:\text{ or }\:x > \frac{9}{2}}$
$\displaystyle B)\;\;20x^33x^217x+6 \:\leq \:0$
We have the cubic function: .$\displaystyle y \:=\:20x^3  3x^2  17x + 6$
. . We want know when this cubic is below the xaxis.
It factors: .$\displaystyle (x + 1)(5x2)(4x3) \:=\;0\quad\Rightarrow\quad x \:=\:1,\:\frac{2}{5},\:\frac{3}{4}$ Code:

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o+oo

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Answer: .$\displaystyle \boxed{x < 1,\;\frac{2}{5} < x < \frac{3}{4}}$
Your answers are correct!
$\displaystyle C)\;\;\frac{x+1}{x3}  \frac{2x^2+3x}{4x^29} \:= \:\frac{5}{2x^29x+9}$
We have: .$\displaystyle \frac{x+1}{x3}  \frac{x(2x+3)}{(2x 3)(2x+3)} \:=\;\frac{5}{(x3)(2x3)}$
. . $\displaystyle = \;\frac{x+1}{x3}  \frac{x}{2x3} \;=\;\frac{5}{(x3)(2x3)} $
Multiply by the LCD, $\displaystyle (x3)(2x3)$
. . $\displaystyle (x+1)(2x3)  x(x3) \:=\:5\quad\Rightarrow\quad x^2 + 2x  8 \:=\:0\quad\Rightarrow\quad(x+4)(x2) \:=\:0$
Therefore: .$\displaystyle \boxed{x \;=\;4,\:2}$
You were right!