The given equation is a quadratic equation. Recall, the standard form of a quadratic equation is expressed as
ax^2 + bx + c = 0
The given equation is
x^2 - 8x = - 128
By adding 128 to both sides of the equation, we have
x^2 - 8x + 128 = - 128 + 128
x^2 - 8x + 128 = 0
By comparing this equation with the standard form equation,
a = 1, b = - 8, c = 128
The formula for solving quadratic equations is expressed as
[tex]\begin{gathered} x\text{ = }\frac{-\text{ b }\pm\sqrt[]{b^2-4ac}}{2a} \\ By\text{ substituting the given values, it becomes} \\ x\text{ = }\frac{-\text{ - 8 }\pm\sqrt[]{-8^2-4(1\times128)}}{2\times1} \\ x\text{ = }\frac{8\pm\sqrt[]{64-512}}{2}\text{ = }\frac{8\pm\sqrt[]{-\text{ 448}}}{2}\text{ = }\frac{8\pm\sqrt[]{-64\text{ }\times\text{ 7}}}{2} \\ x\text{ }=\frac{8\pm(\sqrt[]{-64)}\times\sqrt[]{7}}{2} \\ \text{Note, }\sqrt[]{-\text{ 1}}\text{ = i} \\ \sqrt[]{-64}\text{ = 8i} \\ x\text{ = }\frac{8\pm8i\sqrt[]{7}}{2} \\ \text{Factoring out 2 in the numerator, we have} \\ x\text{ = }\frac{2(4\text{ }\pm4i\sqrt[]{7})}{2} \\ x\text{ = 4 }\pm4i\sqrt[]{7} \end{gathered}[/tex]
Option A is correct
