To find the solutions for the quadratic equation given by
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} \][/tex]
we go through the following steps:
1. Identify the coefficients: For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[
a = 1, \quad b = -2, \quad c = -8
\][/tex]
2. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = (-2)^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36
\][/tex]
3. Find the square root of the discriminant:
[tex]\[
\sqrt{\Delta} = \sqrt{36} = 6
\][/tex]
4. Apply the quadratic formula: The roots [tex]\(x\)[/tex] are given by:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
With [tex]\(b = -2\)[/tex], [tex]\(a = 1\)[/tex], and [tex]\(\sqrt{\Delta} = 6\)[/tex]:
[tex]\[
x = \frac{-(-2) \pm 6}{2 \cdot 1} = \frac{2 \pm 6}{2}
\][/tex]
This gives us two possible solutions:
[tex]\[
x_1 = \frac{2 + 6}{2} = \frac{8}{2} = 4
\][/tex]
[tex]\[
x_2 = \frac{2 - 6}{2} = \frac{-4}{2} = -2
\][/tex]
Therefore, the two solutions for the quadratic equation are:
[tex]\[
x = 4 \quad \text{and} \quad x = -2
\][/tex]
So, the correct answer is:
[tex]\[ x = 4 ; x = -2 \][/tex]
