Solution of the inequality $\frac{2x+3}{4x-5} \geq 0$ is

The fraction $\frac{2x+3}{4x-5}$ is positive or zero when both numerator and denominator have the same sign, or when the numerator is zero. The denominator cannot be zero.

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Finding where the numerator equals zero:

$2x + 3 = 0$

$x = -\frac{3}{2}$

Finding where the denominator equals zero:

$4x - 5 = 0$

$x = \frac{5}{4}$

These critical points divide the number line into three regions.

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Testing the region $x < -\frac{3}{2}$ using $x = -2$:

Numerator: $2(-2) + 3 = -1$ (negative)

Denominator: $4(-2) - 5 = -13$ (negative)

Fraction: $\frac{(-)}{(-)} = (+)$ ✓

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Testing the region $-\frac{3}{2} < x < \frac{5}{4}$ using $x = 0$:

Numerator: $2(0) + 3 = 3$ (positive)

Denominator: $4(0) - 5 = -5$ (negative)

Fraction: $\frac{(+)}{(-)} = (-)$ ✗

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Testing the region $x > \frac{5}{4}$ using $x = 2$:

Numerator: $2(2) + 3 = 7$ (positive)

Denominator: $4(2) - 5 = 3$ (positive)

Fraction: $\frac{(+)}{(+)} = (+)$ ✓

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At $x = -\frac{3}{2}$:

$\frac{2(-\frac{3}{2})+3}{4(-\frac{3}{2})-5} = \frac{0}{-11} = 0$

Since $0 \geq 0$, this point is included.

At $x = \frac{5}{4}$:

The denominator is zero, so this point is excluded.

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The solution is $\left(-\infty, -\frac{3}{2}\right] \cup \left(\frac{5}{4}, \infty\right)$