To solve the quadratic equation [tex]\(2x^2 - 2x - 3 = 0\)[/tex], we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the quadratic equation is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex], so we need to identify [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
a = 2, \quad b = -2, \quad c = -3
\][/tex]
Now we substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}
\][/tex]
Simplify the expression inside the square root:
[tex]\[
x = \frac{2 \pm \sqrt{4 + 24}}{4}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{28}}{4}
\][/tex]
Simplify [tex]\(\sqrt{28}\)[/tex]:
[tex]\[
\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}
\][/tex]
Now substitute back:
[tex]\[
x = \frac{2 \pm 2\sqrt{7}}{4}
\][/tex]
Simplify the fraction:
[tex]\[
x = \frac{2(1 \pm \sqrt{7})}{4} = \frac{1 \pm \sqrt{7}}{2}
\][/tex]
Therefore, the exact solutions are:
[tex]\[
x = \frac{1 - \sqrt{7}}{2} \quad \text{and} \quad x = \frac{1 + \sqrt{7}}{2}
\][/tex]
Next, we calculate the approximate numerical values of these solutions, correct to three decimal places:
[tex]\[
\frac{1 - \sqrt{7}}{2} \approx -0.823
\][/tex]
[tex]\[
\frac{1 + \sqrt{7}}{2} \approx 1.823
\][/tex]
Thus, the solutions to the quadratic equation [tex]\(2x^2 - 2x - 3 = 0\)[/tex] correct to three decimal places are:
[tex]\[
x \approx -0.823 \quad \text{and} \quad x \approx 1.823
\][/tex]
