If the endpoint is (3/2, 1 1/2), and the midpoint is (2/5, -8/5) what is the other endpoint

The answer to your question is  ( -7/10, -4 7/10)

Step-by-step explanation:

Data

Enpoint (3/2, 1 1/2)                  x1 = 3/2     y1 = 1 1/2

Midpoint (2/5, -8/5)                xm = 2/5    ym = -8/5

Entpoint 2 ( x, y)

Formula

Xm = (x1 + x2)/2    Solve for x2    x2 = 2xm - x1

Ym = (y1 + y2)/2     Solve for y2   y2 = 2ym - y1

Substitution

                  x2 = 2(2/5) - 3/2

                  x2 = 4/5 - 3/2

                  x2 = (8 - 15)/10

                 x2 = -7/10

                  y2 = 2(-8/5) - 1 1/2

                 y2 = -16/5 - 3/2

                 y2 = (-32 - 15) / 10

                y2 = -47/10 = -4 7/10

found it xd it ok yt

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