To solve the quadratic equation [tex]\(4x^2 + 4x - 3 = 0\)[/tex], we will use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients from the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
For the equation [tex]\(4x^2 + 4x - 3 = 0\)[/tex], we have:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = 4\)[/tex]
- [tex]\(c = -3\)[/tex]
First, we calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting in our values:
[tex]\[ \Delta = 4^2 - 4(4)(-3) \][/tex]
[tex]\[ \Delta = 16 + 48 \][/tex]
[tex]\[ \Delta = 64 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), we have two real solutions. Now we use the quadratic formula to find these solutions:
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{2 \cdot 4} \][/tex]
[tex]\[ x = \frac{-4 \pm 8}{8} \][/tex]
We have two cases to consider because of the [tex]\(\pm\)[/tex]:
1. When we take the positive square root:
[tex]\[ x = \frac{-4 + 8}{8} \][/tex]
[tex]\[ x = \frac{4}{8} \][/tex]
[tex]\[ x = 0.5 \][/tex]
2. When we take the negative square root:
[tex]\[ x = \frac{-4 - 8}{8} \][/tex]
[tex]\[ x = \frac{-12}{8} \][/tex]
[tex]\[ x = -1.5 \][/tex]
Therefore, the solutions to the equation [tex]\(4x^2 + 4x - 3 = 0\)[/tex] are:
[tex]\[ x = 0.5 \text{ and } x = -1.5 \][/tex]
Thus, the correct answer to the multiple-choice question is:
[tex]\[ \boxed{x = 0.5, -1.5} \][/tex]
