Let

'x' be the first odd integer then it can be written in a form of 2*k - 1, where k is an integer.

In this case the second odd integer can be written as 2*k + 1.

(2k - 1)^2 + 3*(2*k+1) = 24

4k^2 - 2*2*k*1 + 1^2 + 3*2*k +3*1 = 24

4k^2 - 4*k*1 + 1 + 6*k + 3 = 24

4k^2 - 4*k* + 6*k + 1 + 3 = 24

4k^2 - 2*k + 4 = 24

4k^2 + 2*k + 4 - 24 = 0

4k^2 + 2*k - 20 = 0 both sides divide by 2

2k^2 + k - 10 = 0 quadratic equation

Using the formula for solving quadratic equation that is

ax^2 + bx + c = 0 has the following solutions:

x1= (- b + ((b^2 - (4ac))^(1/2))/ (2*a)

x2= (- b - ((b^2 - (4ac))^(1/2))/ (2*a)

For 2k^2 + k - 10 = 0 we will have

k1 = (- 1 + ( 1^2 - 4 * 2 * (-10))^(1/2))/ 2*2

k1 = (- 1 + 81^0.5) / 4

k1 = (- 1 + 9) / 4

k1 = 8/4

k1 = 2 - first solution that is

k1 = (- 1 - ( 1^2 - 4 * 2 * (-10))^(1/2))/ 2*2

k1 = (- 1 - 81^0.5) / 4

k1 = (- 1 - 9) / 4

k1 = -10 /4 - is not a solution since - 10/4 is not integer

Now for the odd numbers we have

2*k1 - 1 = 2*2 - 1 = 3

2*k1 + 1 = 2*2 + 1 = 5

Indeed 3^2 + 3*5 = 9 + 15 = 24

The two consecutive odd integers are 3 and 5.