[tex]\begin{gathered} \text{Given} \\ 4p^2=9p+28 \end{gathered}[/tex]
Equate the given to zero by subtracting both sides by 9p and 28
[tex]\begin{gathered} 4p^2=9p+28 \\ 4p^2-9p-28=9p-9p+28-28 \\ 4p^2-9p-28=\cancel{9p-9p}+\cancel{28-28} \\ 4p^2-9p-28=0 \\ \\ \text{Now it is in the standard form }ax^2+bx+c=0 \\ \text{where} \\ a=4,b=-9,c=-28 \end{gathered}[/tex]
Solve for p using the quadratic equation
[tex]\begin{gathered} p=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ p=\frac{ -(-9) \pm\sqrt{(-9)^2 - 4(4)(-28)}}{ 2(4) } \\ p=\frac{9\pm\sqrt[]{81-(-448)_{}}}{8} \\ p=\frac{9\pm\sqrt[]{81+448_{}}}{8} \\ p=\frac{ 9 \pm\sqrt{529}}{ 8 } \\ p=\frac{ 9 \pm23\, }{ 8 } \end{gathered}[/tex]
Solve for two solutions of p
[tex]\begin{gathered} p=\frac{ 9 \pm23\, }{ 8 } \\ \\ p_1=\frac{9+23}{8} \\ p_1=\frac{32\, }{8} \\ p_1=4 \\ \\ p_2=\frac{9-23}{8} \\ p_2=\frac{-14}{8} \\ p_2=-\frac{7}{4} \\ \\ \text{Therefore, the solutions of the given equation }4p^2=9p+28\text{ are} \\ p=4,\text{ and }p=-\frac{7}{4} \end{gathered}[/tex]
