Answer:
[tex] x = \dfrac{3+\sqrt{29}}{2},\dfrac{3-\sqrt{29}}{2}[/tex]
Step-by-step explanation:
This kinda Questions can be easily done using the Quadratic formula or Shreedhacharya's Formula . Which is used to find the roots of a quadratic equation in Standard form of ax² + bx +c = 0 .
Wrt . Standard form ,
[tex]\implies x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\implies x = \dfrac{-3\pm \sqrt{3^2-4(1)(5)}}{2(1)}\\\\\implies x = \dfrac{-3\pm \sqrt{9+20}}{2} \\\\\implies \boxed{\boxed{ x = \dfrac{3+\sqrt{29}}{2},\dfrac{3-\sqrt{29}}{2}}}[/tex]
This is our required answer.
