The quadratic equation whose one root is $\frac{3+\sqrt{5}}{2-\sqrt{5}}$ is 

The given root is $\alpha =\frac{3+\sqrt{5}}{2-\sqrt{5}}$

Rationalize the denominator

$\begin{array}{rcl}\alpha & =& \frac{3+\sqrt{5}}{2-\sqrt{5}}·\frac{2+\sqrt{5}}{2+\sqrt{5}}\\ & =& \frac{6+3\sqrt{5}+2\sqrt{5}+5}{4-5}\\ & =& \frac{11+5\sqrt{5}}{-1}\\ & =& -11-5\sqrt{5}\end{array}$

SInce the irrational roots are conjugate to each other, the other root is $\beta =-11+5\sqrt{5}$

Use the formula to get the equation having roots $\alpha ,\beta$ as $x^2-x\left(\alpha +\beta \right)-\alpha \beta =0$

Therefore, the equation whose roots are $-11-5\sqrt{5},-11+5\sqrt{5}$ is $x^2-x\left(-22\right)+\left(121-125\right)=0$

Simplifying the above equation will get the required equation as $\boxed{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{22}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0}}$