The sum of three consecutive odd integers is 18 less than 5 times the middle number. find the 3 integers. use an algebraic solution

X +(x+2) + (x+4) = 5(x+2)-18

3x+6 = 5x+10-18

3x+6 = 5x-8

3x=5x-14

-2x=-14

x=7

(x+2) =9

(x+4) =11

7 + 9+ 11 =27

5(9)-18 =45-18 =27

3 numbers are 7, 9 & 11

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15