SOLUTION: please help me find the real solution of this equation {{{(2x+5)^2

SOLUTION: please help me find the real solution of this equation {{{(2x+5)^2 - (2x+5)-6=0}}}

Question 366757: please help me find the real solution of this equation
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
There is a short way and a long way to solve this equation.

The short way requires that we can recognize that this equation is in quadratic form for (2x+5). If this in not clear to you then perhaps using a temporary variable will help:
Let q = (2x+5)
Subsituting this into your equation we get:

This is clearly a quadratic equation. We can solve it by factoring:

Then use the Zero Product property which tells us that this (or any) product can be zero only if one of the factors is zero. So:
q - 3 = 0 or q + 2 = 0
Solving these we get:
q = 3 or q = -2
Of course we want to know what x is not q. So we substitute back in for q:
2x + 5 = 3 or 2x + 5 = -2
Solving these we get:
x = -1 or x = -7/2
After a few of these "quadratic form" equations you will not longer need the temporary variable. You will be able to go straight from

which simplifies to:

then
2x+2 = 0 or 2x+7 = 0
etc.

The long way is to simplify your equation first. Squaring 2x+5 correctly givess us:

Combining like terms we get:

Now we can factor. First the Greatest Common Factor:

Then the trinomial:

Then the Zero Product Property:
2 = 0 or 2x+7 = 0 or x+1 = 0
The first equation has no solution. From the other two equations we get:
x = -7/2 or x = -1 (which are the solutions we got the short way)
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