To find the solutions to the quadratic equation \(10x^2 - 17x + 3 = 0\), we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where \(a\), \(b\), and \(c\) are the coefficients of the equation \(ax^2 + bx + c = 0\). In this case, the coefficients are:
[tex]\[
a = 10, \quad b = -17, \quad c = 3
\][/tex]
First, we need to calculate the discriminant (\(\Delta\)):
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of \(a\), \(b\), and \(c\):
[tex]\[
\Delta = (-17)^2 - 4(10)(3)
\][/tex]
[tex]\[
\Delta = 289 - 120
\][/tex]
[tex]\[
\Delta = 169
\][/tex]
Now that we have the discriminant, we can find the two solutions using the quadratic formula:
[tex]\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a}
\][/tex]
[tex]\[
x_2 = \frac{-b - \sqrt{\Delta}}{2a}
\][/tex]
Substitute \(b = -17\), \(\Delta = 169\), and \(a = 10\):
For \(x_1\):
[tex]\[
x_1 = \frac{17 + \sqrt{169}}{20}
\][/tex]
[tex]\[
x_1 = \frac{17 + 13}{20}
\][/tex]
[tex]\[
x_1 = \frac{30}{20}
\][/tex]
[tex]\[
x_1 = \frac{3}{2} = 1.5
\][/tex]
For \(x_2\):
[tex]\[
x_2 = \frac{17 - \sqrt{169}}{20}
\][/tex]
[tex]\[
x_2 = \frac{17 - 13}{20}
\][/tex]
[tex]\[
x_2 = \frac{4}{20}
\][/tex]
[tex]\[
x_2 = \frac{1}{5} = 0.2
\][/tex]
Thus, the solutions to the quadratic equation \(10x^2 - 17x + 3 = 0\) are \(x = 1.5\) and \(x = 0.2\).
Comparing these results with the provided options, we see:
A. \(x = \frac{3}{2}\) and \(x = \frac{1}{5}\)
Hence, the correct answer is:
[tex]\[
\boxed{\text{A}}
\][/tex]
