Answer:
[tex]h=3, \ h=-5[/tex]
Step-by-step explanation:
[tex](h+1)^2=16[/tex]
[tex](h+1)^2=16[/tex]
[tex]\sqrt{(h+1)^2} = \±\sqrt{16}[/tex] <== you take the square root of both sides
[tex](h+1) = \±\ 4[/tex] <== you take the square root of both sides
[tex](h+1) = +\ 4, \ (h+1) = -\ 4[/tex] <== separate into 2 equations and solve
[tex]h+1 = 4, \ h+1 = -\ 4\\\\. \ -1 -1, \ \ \ \ -1\ \ \ \ -1\\\\h=3, \ h=-5[/tex]
Hope this helps!
