From the quadratic equation formula
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}=-\frac{b}{2a}\pm\frac{\sqrt[]{b^2-4ac}}{2a} \\ \text{where} \\ ax^2+bx+c=0 \end{gathered}[/tex]
Therefore, if x=-4+7i is a solution to the equation; then,
[tex]-4+7i=-\frac{b}{2a}+\frac{\sqrt[]{b^2-4ac}}{2a}[/tex]
Thus, the remaining solution has to be
[tex]-\frac{b}{2a}-\frac{\sqrt[]{b^2-4ac}}{2a}=-4-7i[/tex]
The answer is that the other solution has to be -4-7i, option B
