Given
[tex]x^2+2x-4=0[/tex]
Using the quadratic formula
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Given parameters} \\ a=1 \\ b=2 \\ c=-4 \end{gathered}[/tex]
Subsitute the given parameters into the formula
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ x=\frac{-2\pm\sqrt[]{2^2^{}-4\times1\times-4}}{2\times1} \\ \\ x=\frac{-2\pm\sqrt[]{2^2+16}}{2} \\ \\ x=\frac{-2\pm\sqrt[]{4^{}+16}}{2} \\ x=\frac{-2\pm\sqrt[]{20}}{2} \\ \\ x=\frac{-2\pm\sqrt[]{4\times5}}{2} \\ x=\frac{-2\pm2\sqrt[]{5}}{2} \\ x=-1\pm\sqrt[]{5} \end{gathered}[/tex]
The final answer
Option A and Option D
[tex]\begin{gathered} x=-1-\sqrt[]{5} \\ \text{and} \\ x=-1+\sqrt[]{5} \end{gathered}[/tex]
The summary
