Answer
$$A = \frac{{125}}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Calculate the intersecion points}} \cr
& x = x \cr
& {y^2} - 3y + 12 = - 2{y^2} - 6y + 30 \cr
& 3{y^2} + 3y - 18 = 0 \cr
& {y^2} + y - 6 = 0 \cr
& \left( {y + 3} \right)\left( {y - 2} \right) = 0 \cr
& {y_1} = - 3,\,\,\,{y_2} = 2 \cr
& {\text{The area of the region is given by}} \cr
& A = \int_{ - 3}^2 {\left[ {\left( { - 2{y^2} - 6y + 30} \right) - \left( {{y^2} - 3y + 12} \right)} \right]dy} \cr
& A = \int_{ - 3}^2 {\left( { - 2{y^2} - 6y + 30 - {y^2} + 3y - 12} \right)dy} \cr
& A = \int_{ - 3}^2 {\left( { - 3{y^2} - 3y + 18} \right)dy} \cr
& {\text{Integrating}} \cr
& A = \left[ { - {y^3} - \frac{{3{y^2}}}{2} + 18y} \right]_{ - 3}^2 \cr
& A = \left[ { - {{\left( {2} \right)}^3} - \frac{{3{{\left( {2} \right)}^2}}}{2} + 18\left( {2} \right)} \right] - \left[ { - {{\left( { - 3} \right)}^3} - \frac{{3{{\left( { - 3} \right)}^2}}}{2} + 18\left( { - 3} \right)} \right] \cr
& A = 22 - \left( { - \frac{{81}}{2}} \right) \cr
& A = \frac{{125}}{2} \cr} $$
