To solve the equation [tex]\(12^{x^2 + 5x - 4} = 12^{2x + 6}\)[/tex], we start by recognizing that the bases on both sides are the same. This allows us to set the exponents equal to each other:
[tex]\[ x^2 + 5x - 4 = 2x + 6 \][/tex]
Next, we rearrange the equation to form a standard quadratic equation by moving all terms to one side:
[tex]\[ x^2 + 5x - 4 - 2x - 6 = 0 \][/tex]
Simplify the equation:
[tex]\[ x^2 + 3x - 10 = 0 \][/tex]
Now that we have a quadratic equation, we can solve it using the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( x^2 + 3x - 10 = 0 \)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 3\)[/tex]
- [tex]\(c = -10\)[/tex]
First, calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 3^2 - 4(1)(-10) = 9 + 40 = 49 \][/tex]
Since the discriminant is positive, we will have two real solutions. We now substitute the values into the quadratic formula:
[tex]\[ x = \frac{-3 \pm \sqrt{49}}{2(1)} = \frac{-3 \pm 7}{2} \][/tex]
This gives us two solutions:
1. [tex]\( x_1 = \frac{-3 + 7}{2} = \frac{4}{2} = 2 \)[/tex]
2. [tex]\( x_2 = \frac{-3 - 7}{2} = \frac{-10}{2} = -5 \)[/tex]
Therefore, the solutions to the equation are:
[tex]\[ x = 2 \quad \text{and} \quad x = -5 \][/tex]
So, the correct answers are [tex]\( x = 2 \)[/tex] and [tex]\( x = -5 \)[/tex].
