We need to complete some squares.
Rewrite: 7(x^2+2x+1)+2(y^2-4y+4)-15+1=0⇒7(x+1)^2+2(y-2)^2=14.
Divide through by 14: (x+1)^2/2+(y-2)^2/7=1, which is the standard form of an ellipse with centre (-1,2) and minor and major axes lengths 2sqrt(2) and 2sqrt(7). To draw the ellipse mark its centre at (-1,2), and draw the axes of the ellipse using this point as a secondary origin for the graph. On these secondary axes, which need to be calibrated just like the normal x and y axes are, mark the distance sqrt(2) on either side of the secondary x-axis and sqrt(7) on either side of the secondary y-axis. The square roots of 2 and 7 are 1.414 and 2.646 respectively. You should now be able to draw an ellipse passing through these secondary points. You will be able to see that on the normal axes the location of the intercepts on the secondary axes are (-2.414,2) and (0.414,2) for x, and (-1,4.646) and (-1,-0.646) for y. You can also plot the normal x and y intercepts by substituting y and x equal to zero respectively. When x=0, the y intercepts are y=2+sqrt(7/2)=3.871 and 0.129; and when y=0, the x intercepts are x=+sqrt(3/7)-1=-0.345 and -1.655.
