To solve the equation \((x+2)^2 + 12(x+2) - 14 = 0\) using \(u\) substitution and the quadratic formula, let's follow these steps:
1. Substitution:
Introduce a substitution \( u = x + 2 \). This simplifies the equation because it reduces the complexity. Substitute \( u \) into the equation:
[tex]\[
(x+2)^2 + 12(x+2) - 14 = 0
\][/tex]
becomes:
[tex]\[
u^2 + 12u - 14 = 0
\][/tex]
2. Quadratic Formula:
The quadratic formula states that for any quadratic equation of the form \( au^2 + bu + c = 0 \), the solutions can be found using:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, \( a = 1 \), \( b = 12 \), and \( c = -14 \). Substitute these values into the quadratic formula to find \( u \):
[tex]\[
u = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1}
\][/tex]
Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot (-14) = 144 + 56 = 200
\][/tex]
Now, solve for \( u \):
[tex]\[
u = \frac{-12 \pm \sqrt{200}}{2}
\][/tex]
Since \(\sqrt{200} = 10\sqrt{2}\),
[tex]\[
u = \frac{-12 \pm 10\sqrt{2}}{2} = -6 \pm 5\sqrt{2}
\][/tex]
Thus, the solutions for \( u \) are:
[tex]\[
u_1 = -6 + 5\sqrt{2}, \quad u_2 = -6 - 5\sqrt{2}
\][/tex]
3. Back Substitution:
Recall that \( u = x + 2 \). We need to solve for \( x \):
[tex]\[
u_1 = x + 2 \implies x_1 = u_1 - 2
\][/tex]
[tex]\[
u_2 = x + 2 \implies x_2 = u_2 - 2
\][/tex]
Substitute the values of \( u_1 \) and \( u_2 \) back:
[tex]\[
x_1 = -6 + 5\sqrt{2} - 2 = -8 + 5\sqrt{2}
\][/tex]
[tex]\[
x_2 = -6 - 5\sqrt{2} - 2 = -8 - 5\sqrt{2}
\][/tex]
Therefore, the solutions are:
[tex]\[
x = -8 \pm 5\sqrt{2}
\][/tex]
Hence, after all the steps, the solutions to the equation \((x+2)^2 + 12(x+2) - 14 = 0\) are:
[tex]\[
x = -8 \pm 5\sqrt{2}
\][/tex]
So, the correct answer is:
[tex]\[
x = -8 \pm 5\sqrt{2}
\][/tex]
