(y+10)^2-72=0 solve and simplify where y is a real number?

Question

(y+10)^2-72=0

so far:
(y^2+20y+100)-72=0
y^2+20y+100-72=0
y^2+20*y+28=0

now, to obtain a real number
to plug in, if there is more than
one solution I am supposed to separate them
with a comma

Answer 1

(y+10)^2-72=0
(y+10)^2 = 72
y+10 = +- sqrt(72)
y= -10 +- sqrt(72)
-10 + sqrt(72), -10 - sqrt(72)
s

Answer 2

(y+10)^2-72 = 0
(y+10)^2 = 72
(y+10) = sqrt72
y = sqrt(9*8) - 10
y = 3sqrt8 - 10
y = -1.51
y = -18.48

Answer 3

Use the quadratic formula where a=1, b=20 and c=28

(-b+(or -) sqrt(b^2 - 4ac))/2a

This will give you two answers. When you use the plus after the -b it will give you one answer and when you use the - after the -b, it will give you another answer.

Answer 4

you can use Quartic equation to solve the equation.
y1 = -6*(2)^(1/2) - 10=-18.48528
y2 = 6*(2)^(1/2) -10 =-1.514719

Answer 5

y+10 = sqrt(72)
so y = sqrt(72) - 10
so y = 8.485 - 10 or y = -8.485 - 10
so y = -1.515 or y = -18.485 (3 decimal places)

Answer 6

Yes you're supposed to state both answers as y=-18.48, y=-1.51