To solve the quadratic equation [tex]\(0 = x^2 - 2x - 3\)[/tex] using the quadratic formula, we need to correctly substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula. The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are taken from the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
For the equation [tex]\(0 = x^2 - 2x - 3\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -2\)[/tex]
- [tex]\(c = -3\)[/tex]
We will now substitute these values into the quadratic formula step-by-step:
1. Put the value of [tex]\(b\)[/tex] into the formula:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}
\][/tex]
2. Simplify inside the square root and fraction:
[tex]\[
x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{4 + 12}}{2}
\][/tex]
3. Combine the terms inside the square root:
[tex]\[
x = \frac{2 \pm \sqrt{16}}{2}
\][/tex]
4. Taking the square root of 16:
[tex]\[
x = \frac{2 \pm 4}{2}
\][/tex]
Given the original options, the ones that match this process are:
[tex]\[
\frac{-2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}
\][/tex]
and
[tex]\[
\frac{-2 \pm \sqrt{-2^2 - 4(1)(-3)}}{2(1)}
\][/tex]
Among these choices, while they appear similar, only the following shows the fully simplified version and correct signs:
[tex]\[
\frac{-2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}
\][/tex]
Thus, the correct substitution of the values [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic formula is:
[tex]\[
\boxed{\frac{-2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}}
\][/tex]
