$\begin{align}  & \text{Given equation }7+12x-2{{x}^{2}} \\ & a=-2,b=12,\text{ }c=7 \\ & \text{Sum of roots (}\alpha +\beta )=-\frac{b}{a}=-\frac{12}{-2}=6 \\ & \text{product of roots }=(\alpha \beta )=\frac{c}{a}=-\frac{7}{2} \\ & \text{For the equation whose roots are }\frac{\alpha }{\beta }\text{ and }\frac{\beta }{\alpha } \\ & \text{Sum of new roots }=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta } \\ & \text{Sum of new roots}=\frac{{{6}^{2}}-2(-\tfrac{7}{2})}{(-\tfrac{7}{2})}=\frac{36+7}{-\tfrac{7}{2}}=43\times \frac{-2}{7}=\frac{86}{7} \\ & \text{Product  of new roots=}\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }=1 \\ & \text{The new eqaution will be}=\text{ }{{x}^{2}}-\left( \frac{\alpha }{\beta }+\frac{\beta }{\alpha } \right)x+\left( \frac{\alpha }{\beta }\times \frac{\beta }{\alpha } \right)=0 \\ & ={{x}^{2}}+\frac{86}{7}x+1=0 \\ & =7{{x}^{2}}+86x+7=0 \\\end{align}$