If $a^2 +b^2 + 2b + 4a + 5 = 0$, then the value of $\frac{2a-3b}{2a+3b}$ is equal to:

$\frac{1}{7}$

If $a^2 +b^2 + 2b + 4a + 5 = 0$

We know that,

(a + b)²  = a² + b² + 2ab

 a² + b² + 2b + 4a + 5 = 0,

= a² + 4a + b² + 2b + 5 = 0

= a² + 4a + 4 + b² + 2b + 1 = 0

= (a + 2)² + (b + 1)² = 0

So, a + 2 = 0

= a = -2

And, b + 1 = 0

= b = -1

Put these values in the desired equation,

 $\frac{2a-3b}{2a+3b}$ = $\frac{2(-2)-3(-1)}{2(-2)+3(-1)}$ = \(\frac{1}{7}\)