To solve the quadratic equation \( x^2 - 0.14x + 0.0049 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -0.14 \), and \( c = 0.0049 \).
First, let's calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (-0.14)^2 = 0.0196 \] \[ 4ac = 4 \cdot 1 \cdot 0.0049 = 0.0196 \] \[ b^2 - 4ac = 0.0196 - 0.0196 = 0 \]
Since the discriminant is zero, there is one real double root.
Now we can use the quadratic formula to find the root:
\[ x = \frac{-(-0.14) \pm \sqrt{0}}{2 \cdot 1} = \frac{0.14 \pm 0}{2} = \frac{0.14}{2} = 0.07 \]
Thus, the solution to the quadratic equation \( x^2 - 0.14x + 0.0049 = 0 \) is:
\[ \boxed{0.07} \]
5 answers