To solve the equation [tex]\((2x + 3)^2 + 8(2x + 3) + 11 = 0\)[/tex], we will use [tex]\(u\)[/tex]-substitution and the quadratic formula.
### Step 1: Substitution
Let [tex]\(u = 2x + 3\)[/tex]. Then the equation transforms into:
[tex]\[ u^2 + 8u + 11 = 0 \][/tex]
### Step 2: Solving the Quadratic Equation
The equation [tex]\(u^2 + 8u + 11 = 0\)[/tex] is a standard quadratic form [tex]\(au^2 + bu + c = 0\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 11\)[/tex].
To find the solutions for [tex]\(u\)[/tex], we use the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ u = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 11}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-8 \pm \sqrt{64 - 44}}{2} \][/tex]
[tex]\[ u = \frac{-8 \pm \sqrt{20}}{2} \][/tex]
[tex]\[ u = \frac{-8 \pm 2\sqrt{5}}{2} \][/tex]
[tex]\[ u = -4 \pm \sqrt{5} \][/tex]
Thus, the solutions for [tex]\(u\)[/tex] are:
[tex]\[ u_1 = -4 + \sqrt{5} \][/tex]
[tex]\[ u_2 = -4 - \sqrt{5} \][/tex]
### Step 3: Converting Back to [tex]\(x\)[/tex]
Recall that [tex]\(u = 2x + 3\)[/tex]. So, we substitute back to find [tex]\(x\)[/tex]:
For [tex]\(u_1\)[/tex]:
[tex]\[ -4 + \sqrt{5} = 2x + 3 \][/tex]
[tex]\[ 2x = -4 + \sqrt{5} - 3 \][/tex]
[tex]\[ 2x = -7 + \sqrt{5} \][/tex]
[tex]\[ x = \frac{-7 + \sqrt{5}}{2} \][/tex]
For [tex]\(u_2\)[/tex]:
[tex]\[ -4 - \sqrt{5} = 2x + 3 \][/tex]
[tex]\[ 2x = -4 - \sqrt{5} - 3 \][/tex]
[tex]\[ 2x = -7 - \sqrt{5} \][/tex]
[tex]\[ x = \frac{-7 - \sqrt{5}}{2} \][/tex]
Thus, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x_1 = \frac{-7 + \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{-7 - \sqrt{5}}{2} \][/tex]
### Conclusion
The solutions to the equation [tex]\((2x + 3)^2 + 8(2x + 3) + 11 = 0\)[/tex] are:
[tex]\[ x = \frac{-7 \pm \sqrt{5}}{2} \][/tex]
