Question

Solve the equation using quadratic formula.

$2p^2-7p+3=0$

Short answer

The solutions are$p=\frac{1}{2},3$.

Step-by-step solution

 Given Information. 

The given equation is $2p^2-7p+3=0$.

Write the quadratic equation in standard form and identify a,b,c.

The quadratic equation is in standard form.

So,

$a=2,b=-7,c=3$.

Substitute the values in quadratic formula. 

$$\begin{gathered} p=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ p=\frac{-(-7)\pm \sqrt{(-7{)}^2-4(2\left)\right(3)}}{2\left(2\right)} \\ p=\frac{7\pm \sqrt{49-24}}{4} \end{gathered}$$

Simplify the square root.

$$\begin{gathered} p=\frac{7\pm \sqrt{25}}{4} \\ p=\frac{7\pm 5}{4} \end{gathered}$$

Rewrite to show two solutions. 

$$\begin{gathered} p_1=\frac{7+5}{4} \\ {p}_1=\frac{12}{4}=3, \\ {p}_2=\frac{7-5}{4} \\ {p}_2=\frac{2}{4}=\frac{1}{2} \end{gathered}$$

Thus, the solutions are$p=\frac{1}{2},3$.