To solve the quadratic equation (x^2 - 12x + 0.35 = 0), we can use the quadratic formula:
[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
]
In this case, the coefficients are:
- (a = 1)
- (b = -12)
- (c = 0.35)
Now, we can plug these values into the formula:
- Calculate the discriminant (b^2 - 4ac):
[
b^2 = (-12)^2 = 144
]
[
4ac = 4 \cdot 1 \cdot 0.35 = 1.4
]
[
b^2 - 4ac = 144 - 1.4 = 142.6
]
- Now substitute into the quadratic formula:
[
x = \frac{-(-12) \pm \sqrt{142.6}}{2 \cdot 1}
]
[
x = \frac{12 \pm \sqrt{142.6}}{2}
]
- Calculate (\sqrt{142.6}):
[
\sqrt{142.6} \approx 11.92 \text{ (approximately)}
]
- Substitute back into the equation:
[
x = \frac{12 \pm 11.92}{2}
]
This gives us two solutions:
- (x_1 = \frac{12 + 11.92}{2} = \frac{23.92}{2} \approx 11.96)
- (x_2 = \frac{12 - 11.92}{2} = \frac{0.08}{2} \approx 0.04)
Therefore, the solutions to the equation (x^2 - 12x + 0.35 = 0) are approximately:
[
x \approx 11.96 \quad \text{and} \quad x \approx 0.04
]
