Answer :
Given the demand function :
[tex]p=\frac{28}{x+1}[/tex]And the supply function:
[tex]p=1+0.2x[/tex]So,
[tex]\begin{gathered} \frac{28}{x+1}=1+0.2x \\ \\ (1+0.2x)\cdot(x+1)=28 \\ 1\cdot(x+1)+0.2x\cdot(x+1)=28 \\ x+1+0.2x^2+0.2x=28 \\ 0.2x^2+1.2x+1=28 \\ 0.2x^2+1.2x-27=0 \end{gathered}[/tex]using the general rule to solve the quadratic equation:
a = 0.2 , b = 1.2 , c = -27
so,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}=\frac{-1.2\pm\sqrt[]{1.2^2-4\cdot0.2\cdot(-27)}}{2\cdot0.2}[/tex][tex]\begin{gathered} x=\frac{-1.2\pm\sqrt[]{23.04}}{0.4} \\ \\ x=\frac{-1.2\pm4.8}{0.4} \\ \\ x=\frac{-1.2+4.8}{0.4}=\frac{3.6}{0.4}=9 \\ OR \\ x=\frac{-1.2-4.8}{0.4}=-15 \end{gathered}[/tex]So, the answer is x = 9