4х²+12х+9=(х+4)²

To solve the equation \( 4x^2 + 12x + 9 = (x + 4)^2 \), we first need to simplify the right side of the equation.

We can expand \((x + 4)^2\):

\[ (x + 4)^2 = x^2 + 8x + 16 \]

Now, we can rewrite the equation:

\[ 4x^2 + 12x + 9 = x^2 + 8x + 16 \]

Next, we move all terms to one side of the equation to set it equal to zero:

\[ 4x^2 + 12x + 9 - x^2 - 8x - 16 = 0 \]

Combine like terms:

\[ (4x^2 - x^2) + (12x - 8x) + (9 - 16) = 0 \]

This simplifies to:

\[ 3x^2 + 4x - 7 = 0 \]

Now, we can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 3\), \(b = 4\), and \(c = -7\). We will calculate the discriminant:

\[ b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot (-7) = 16 + 84 = 100 \]

Now we can substitute back into the quadratic formula:

\[ x = \frac{-4 \pm \sqrt{100}}{2 \cdot 3} \] \[ x = \frac{-4 \pm 10}{6} \]

This gives us two potential solutions:

1. \(x = \frac{-4 + 10}{6} = \frac{6}{6} = 1\)

2. \(x = \frac{-4 - 10}{6} = \frac{-14}{6} = \frac{-7}{3}\)

Thus, the solutions to the equation \(4x^2 + 12x + 9 = (x + 4)^2\) are:

\[ x = 1 \quad \text{and} \quad x = -\frac{7}{3} \]