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To solve the quadratic equation [tex]\(x^2 + 6x - 10 = 0\)[/tex] using the quadratic formula, we will follow these steps:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = -10\)[/tex]
### Step 1: Calculate the Discriminant
The discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the given values:
[tex]\[ \Delta = (6)^2 - 4(1)(-10) = 36 + 40 = 76 \][/tex]
Since the discriminant is positive ([tex]\(\Delta = 76\)[/tex]), we have two distinct real solutions.
### Step 2: Apply the Quadratic Formula
We now substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex] into the quadratic formula:
[tex]\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-6 \pm \sqrt{76}}{2(1)} \][/tex]
### Step 3: Calculate Each Solution
The solutions are:
[tex]\[ x_1 = \frac{-6 + \sqrt{76}}{2} \quad \text{and} \quad x_2 = \frac{-6 - \sqrt{76}}{2} \][/tex]
### Step 4: Simplify the Solutions
Simplifying the expressions, we get:
[tex]\[ x_1 = \frac{-6 + \sqrt{76}}{2} = \frac{-6}{2} + \frac{\sqrt{76}}{2} = -3 + \sqrt{19} \][/tex]
[tex]\[ x_2 = \frac{-6 - \sqrt{76}}{2} = \frac{-6}{2} - \frac{\sqrt{76}}{2} = -3 - \sqrt{19} \][/tex]
Thus, the exact solutions to the equation [tex]\(x^2 + 6x - 10 = 0\)[/tex] are:
[tex]\[ x = -3 + \sqrt{19}, \; -3 - \sqrt{19} \][/tex]
### Conclusion
The correct choice is:
A. [tex]\(x = -3 + \sqrt{19}, -3 - \sqrt{19}\)[/tex]
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = -10\)[/tex]
### Step 1: Calculate the Discriminant
The discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the given values:
[tex]\[ \Delta = (6)^2 - 4(1)(-10) = 36 + 40 = 76 \][/tex]
Since the discriminant is positive ([tex]\(\Delta = 76\)[/tex]), we have two distinct real solutions.
### Step 2: Apply the Quadratic Formula
We now substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex] into the quadratic formula:
[tex]\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-6 \pm \sqrt{76}}{2(1)} \][/tex]
### Step 3: Calculate Each Solution
The solutions are:
[tex]\[ x_1 = \frac{-6 + \sqrt{76}}{2} \quad \text{and} \quad x_2 = \frac{-6 - \sqrt{76}}{2} \][/tex]
### Step 4: Simplify the Solutions
Simplifying the expressions, we get:
[tex]\[ x_1 = \frac{-6 + \sqrt{76}}{2} = \frac{-6}{2} + \frac{\sqrt{76}}{2} = -3 + \sqrt{19} \][/tex]
[tex]\[ x_2 = \frac{-6 - \sqrt{76}}{2} = \frac{-6}{2} - \frac{\sqrt{76}}{2} = -3 - \sqrt{19} \][/tex]
Thus, the exact solutions to the equation [tex]\(x^2 + 6x - 10 = 0\)[/tex] are:
[tex]\[ x = -3 + \sqrt{19}, \; -3 - \sqrt{19} \][/tex]
### Conclusion
The correct choice is:
A. [tex]\(x = -3 + \sqrt{19}, -3 - \sqrt{19}\)[/tex]
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