Question

Solve the quadratic function and simplify if needed of course pls 4p2=9p+28

Answer 1

`\begin{gathered} \text{Given} \\ 4p^2=9p+28 \end{gathered}`

Equate the given to zero by subtracting both sides by 9p and 28

`\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}`

Solve for p using the quadratic equation

`\begin{gathered} p=( -b \pm√(b^2 - 4ac))/( 2a ) \\ p=( -(-9) \pm√((-9)^2 - 4(4)(-28)))/( 2(4) ) \\ p=\frac{9\pm\sqrt[]{81-(-448)_{}}}{8} \\ p=\frac{9\pm\sqrt[]{81+448_{}}}{8} \\ p=( 9 \pm√(529))/( 8 ) \\ p=( 9 \pm23\, )/( 8 ) \end{gathered}`

Solve for two solutions of p

`\begin{gathered} p=( 9 \pm23\, )/( 8 ) \\ \\ p_1=(9+23)/(8) \\ p_1=(32\, )/(8) \\ p_1=4 \\ \\ p_2=(9-23)/(8) \\ p_2=(-14)/(8) \\ p_2=-(7)/(4) \\ \\ \text{Therefore, the solutions of the given equation }4p^2=9p+28\text{ are} \\ p=4,\text{ and }p=-(7)/(4) \end{gathered}`