To solve the quadratic equation \(3x^2 + x - 5 = 0\), we will use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients \(a\), \(b\), and \(c\) are:
\(a = 3\), \(b = 1\), and \(c = -5\).
First, we need to calculate the discriminant (\(\Delta\)), which is part of the quadratic formula under the square root. The discriminant is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of \(a\), \(b\), and \(c\):
[tex]\[
\Delta = (1)^2 - 4(3)(-5)
\][/tex]
[tex]\[
\Delta = 1 - (-60)
\][/tex]
[tex]\[
\Delta = 1 + 60
\][/tex]
[tex]\[
\Delta = 61
\][/tex]
The discriminant is 61. Since the discriminant is positive, we will have two distinct real roots.
Next, we calculate the roots using the quadratic formula. The two possible solutions are:
[tex]\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a} \quad \text{and} \quad x_2 = \frac{-b - \sqrt{\Delta}}{2a}
\][/tex]
Substituting the values of \(a\), \(b\), and \(\Delta\):
[tex]\[
x_1 = \frac{-1 + \sqrt{61}}{6}
\][/tex]
[tex]\[
x_2 = \frac{-1 - \sqrt{61}}{6}
\][/tex]
To find the approximate values of these roots, we will compute the value inside the square root and then evaluate the expressions:
[tex]\[
\sqrt{61} \approx 7.81
\][/tex]
Now substituting this approximation into the expressions for \(x_1\) and \(x_2\):
[tex]\[
x_1 = \frac{-1 + 7.81}{6} \approx \frac{6.81}{6} \approx 1.14
\][/tex]
[tex]\[
x_2 = \frac{-1 - 7.81}{6} \approx \frac{-8.81}{6} \approx -1.47
\][/tex]
Thus, the solutions to the quadratic equation \(3x^2 + x - 5 = 0\) are approximately:
[tex]\[
x_1 \approx 1.14 \quad \text{and} \quad x_2 \approx -1.47
\][/tex]
