Given equation
[tex]x^2-5x+2=0[/tex]
use the quadratic equation,
[tex]x_{}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex][tex]a=1,\: b=-5,\: c=2[/tex][tex]x_{}=\frac{-\left(-5\right)\pm\sqrt{\left(-5\right)^2-4\cdot\:1\cdot\:2}}{2\cdot\:1}[/tex]
consider the discrement,
[tex]\begin{gathered} \sqrt{\left(-5\right)^2-4\cdot\:1\cdot\:2} \\ =\sqrt{5^2-4\cdot\:1\cdot\:2} \\ =\sqrt{5^2-8} \\ =\sqrt{25-8} \\ =\sqrt{17} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} x_{}=\frac{-\left(-5\right)\pm\sqrt{17}}{2\cdot\:1} \\ x_1=\frac{-\left(-5\right)+\sqrt{17}}{2\cdot\:1},\: x_2=\frac{-\left(-5\right)-\sqrt{17}}{2\cdot\:1} \\ \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-\left(-5\right)+\sqrt{17}}{2\cdot\:1} \\ =\frac{5+\sqrt{17}}{2\cdot\:1} \\ x=\frac{5+\sqrt{17}}{2} \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-\left(-5\right)-\sqrt{17}}{2\cdot\:1} \\ =\frac{5-\sqrt{17}}{2\cdot\:1} \\ =\frac{5-\sqrt{17}}{2} \\ \end{gathered}[/tex]
Answer : The solutions are, option A and D
[tex]\begin{gathered} x=\frac{5+\sqrt{17}}{2},\: x=\frac{5-\sqrt{17}}{2} \\ (or) \\ x=\frac{5\pm\sqrt[]{17}}{2} \end{gathered}[/tex]
