Find three consecutive odd integers such that the product of the second and the third integers is twenty-six more than three times the first integer.

Step-by-step explanation:

first number=x

second number=x+2

third number=x+4

(x+2)(x+4)=3x+26

x(x+4)+2(x+4)=3x+26

x²+4x+2x+8=3x+26

x²+6x+8=3x+26

x²+6x-3x=26-8

x²+3x=18

x²+3x-18=0

you can use that to find the numbers

6, 8 and 10

Step-by-step explanation:

Let the 3 consecutive odd numbers be x, x+2, x+4

product of the second and the third integer

(x+2)(x+4)

twenty-six more than three times the first integer is expressed as;

3x + 26

Equate

(x+2)(x+4) = 3x + 26

x²+4x+2x+8 = 3x + 26

x²+6x +8= 3x+26

x²+3x +8-26 = 0

x²+3x-18 = 0

Factorize

x²-6x+3x-18 = 0

x(x-6)+3(x-6) = 0

x+3 = 0 and x-6 = 0

x = -3 and 6

The first 3 number is are 6, 8 and 10

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation:

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.