Answer: The required solution is
[tex]x=1.19,~-4.19.[/tex]
Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square :
[tex]x^2+3x-5=0.[/tex]
We have
[tex]x^2+3x-5=0\\\\\Rightarrow x^2+3x=5\\\\\Rightarrow x^2+2\times x\times \dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2=5+\left(\dfrac{3}{2}\right)^2\\\\\\\Rightarrow \left(x+\dfrac{3}{2}\right)^2=5+\dfrac{9}{4}\\\\\\\Rightarrow x+\dfrac{3}{2}=\pm\dfrac{\sqrt{29}}{2}\\\\\\\Rightarrow x=-\dfrac{3}{2}\pm\dfrac{\sqrt{29}}{2}\\\\\\\Rightarrow x=\dfrac{-3\pm\sqrt{29}}{2}\\\\\Rightarrow x=\dfrac{-3\pm5.38}{2}\\\\\\\Rightarrow x=\dfrac{2.38}{2},~~\dfrac{-8.38}{2}\\\\\\\Rightarrow x=1.19,~-4.19[/tex]
Thus, the required solution is
[tex]x=1.19,~-4.19.[/tex]
