Question 5221
I have a hunch that you forgot an 'x' and meant

{{{(2x+5)^2-(2x+5)-6=0}}}

use y instead of 2x+5, to get

{{{y^2-y-6 = 0}}}

solve it for y:


*[invoke quadratic "y", 1, -1, -6]

y=3, y=-2, or, knowing that y=2x+5, you have two equations

2x+5 = 3, or 2x=-2, or x = -1
2x+5 = -2, or 2x=-7, or x = -3.5.

Double check my answer please.

Answer: x=-1, x=-3.5