To find the roots of the quadratic equation [tex]\(2x^2 + 7x - 3 = 0\)[/tex], we will use the quadratic formula. The quadratic formula for an equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 7\)[/tex]
- [tex]\(c = -3\)[/tex]
First, we calculate the discriminant [tex]\(\Delta\)[/tex], which is:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the discriminant formula:
[tex]\[
\Delta = 7^2 - 4 \cdot 2 \cdot (-3) = 49 + 24 = 73
\][/tex]
Next, we use the quadratic formula to find the roots:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substitute the values of [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[
x = \frac{-7 \pm \sqrt{73}}{2 \cdot 2} = \frac{-7 \pm \sqrt{73}}{4}
\][/tex]
This results in two roots:
[tex]\[
x_1 = \frac{-7 + \sqrt{73}}{4}
\][/tex]
[tex]\[
x_2 = \frac{-7 - \sqrt{73}}{4}
\][/tex]
Hence, the roots of the quadratic equation [tex]\(2x^2 + 7x - 3 = 0\)[/tex] are:
[tex]\[
x_1 = \frac{-7 + \sqrt{73}}{4}
\][/tex]
[tex]\[
x_2 = \frac{-7 - \sqrt{73}}{4}
\][/tex]
