The equation given is,
[tex]x^2+3x-5=0[/tex]
Simplify
[tex]\begin{gathered} Solve\text{ with the quadratic formula} \\ x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1} \\ x_{1,\:2}=\frac{-3\pm \sqrt{29}}{2\cdot \:1} \\ \mathrm{Separate\:the\:solutions} \\ x_1=\frac{-3+\sqrt{29}}{2\cdot \:1},\:x_2=\frac{-3-\sqrt{29}}{2\cdot \:1} \\ x_1=\frac{-3+\sqrt{29}}{2},\:x_2=\frac{-3-\sqrt{29}}{2} \\ The\:solutions\:to\:the\:quadratic\:equation\:are: \\ x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2} \end{gathered}[/tex]
Hence, Option C and D are the correct answers.
