Solve the equation
`10^((x+1)(3x+4))-2*10^((x+1)(x+2))=10^(1-x-x^(2)).`

Answer

To solve the equation

\[ 10^{(x+1)(3x+4)} - 2 \cdot 10^{(x+1)(x+2)} = 10^{(1-x-x^2)}, \]

we will follow these steps:

### Step 1: Simplify the Exponents
First, we simplify the exponents in the equation. We can expand the terms in the exponents:

\[ (x+1)(3x+4) = 3x^2 + 7x + 4, \]
\[ (x+1)(x+2) = x^2 + 3x + 2. \]

Thus, our equation becomes:

\[ 10^{3x^2 + 7x + 4} - 2 \cdot 10^{x^2 + 3x + 2} = 10^{1 - x - x^2}. \]

### Step 2: Rewrite the Equation
We can rewrite the equation as:

\[ 10^{3x^2 + 7x + 4} - 2 \cdot 10^{x^2 + 3x + 2} = 10^{1 - x - x^2}. \]

### Step 3: Multiply by \(10^{-(x^2 + 3x + 2)}\)
To simplify further, we can multiply both sides by \(10^{-(x^2 + 3x + 2)}\):

\[ 10^{(3x^2 + 7x + 4) - (x^2 + 3x + 2)} - 2 = 10^{(1 - x - x^2) - (x^2 + 3x + 2)}. \]

This simplifies to:

\[ 10^{2x^2 + 4x + 2} - 2 = 10^{-2x - x^2 - 1}. \]

### Step 4: Set \(y = 10^{x^2 + 2x}\)
Let \(y = 10^{x^2 + 2x}\). Therefore, we can rewrite the equation as:

\[ y^2 - 2 = \frac{10^{-1}}{y}. \]

Multiplying both sides by \(y\) gives:

\[ y^3 - 2y + 1 = 0. \]

### Step 5: Solve the Cubic Equation
Now we need to solve the cubic equation \(y^3 - 2y + 1 = 0\). We can use the Rational Root Theorem or synthetic division to find possible rational roots.

Testing \(y = 1\):

\[ 1^3 - 2 \cdot 1 + 1 = 0. \]

So, \(y = 1\) is a root. We can factor the cubic equation as:

\[ (y - 1)(y^2 + y - 1) = 0. \]

### Step 6: Solve the Quadratic Equation
Now we solve the quadratic equation \(y^2 + y - 1 = 0\) using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}. \]

### Step 7: Find Positive Values of \(y\)
The possible values for \(y\) are:

1. \(y = 1\)
2. \(y = \frac{-1 + \sqrt{5}}{2}\) (positive)
3. \(y = \frac{-1 - \sqrt{5}}{2}\) (negative, discard)

### Step 8: Back Substitute for \(x\)
Now we back substitute for \(x\):

1. For \(y = 1\):
\[ 10^{x^2 + 2x} = 1 \implies x^2 + 2x = 0 \implies x(x + 2) = 0 \implies x = 0 \text{ or } x = -2. \]

2. For \(y = \frac{-1 + \sqrt{5}}{2}\):
\[ 10^{x^2 + 2x} = \frac{-1 + \sqrt{5}}{2} \implies x^2 + 2x = \log_{10}\left(\frac{-1 + \sqrt{5}}{2}\right). \]

### Step 9: Solve for \(x\)
For \(x^2 + 2x - \log_{10}\left(\frac{-1 + \sqrt{5}}{2}\right) = 0\), we use the quadratic formula again:

\[ x = \frac{-2 \pm \sqrt{4 + 4\log_{10}\left(\frac{-1 + \sqrt{5}}{2}\right)}}{2} = -1 \pm \sqrt{1 + \log_{10}\left(\frac{-1 + \sqrt{5}}{2}\right)}. \]

### Final Solutions
Thus, the solutions to the original equation are:

\[ x = 0, \quad x = -2, \quad x = -1 \pm \sqrt{1 + \log_{10}\left(\frac{-1 + \sqrt{5}}{2}\right)}. \]