Find three consecutive even integers such that their sum decreased by 34 is equal to the first integer. Find the first integer.

The first number: 14

The second number: 16

The third number: 18

Step-by-step explanation:

Equations

Let's assume:

x=the first even number.

x+2=the second even number

x+4=the third even number

The sum of all three is: x + x + 2 + x + 4

The condition states the sum decreased by 34 equals the first number:

x + x + 2 + x + 4 - 34 = x

Moving all the variables to the left side and the numbers to the right side:

x + x + 2 + x + 4 - 34 - x = 34 - 2 - 4

Simplifying:

2x =  28

Solving:

x = 28 / 2 = 14

The first number: 14

The second number: 16

The third number: 18

∞ this is your sign

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