How do you solve \[4{{p}^{2}}=-7p-3~\] using the quadratic formula?

Hint: The problem that we are given is based on a quadratic equation that must be solved using the quadratic formula, also known as Sridhar Acharya’s formula for solving quadratic equations. A quadratic equation is basically an equation of the second degree, meaning it contains at least one term that is squared. We solve it using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Here, $a$ , $b$ and $c$ represents the coefficient of ${{x}^{2}}$ , $x$ and $1$ respectively. Thus, putting the values accordingly in the formula and considering both the signs of the root square we get the solutions of the quadratic equation.

Complete step by step solution:
We are given the equation
 \[4{{p}^{2}}=-7p-3~\]
We add both the sides of the above equation with $\left( 7p+3 \right)$ and get
\[\Rightarrow 4{{p}^{2}}+7p+3=-7p-3~+7p+3\]
\[\Rightarrow 4{{p}^{2}}+7p+3=0\]
This is a quadratic equation.
A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is \[ax{}^\text{2}+bx+c=0\] with $a$ , $b$ , and $c$ being constants, or numerical coefficients, and $x$ is an unknown variable. One absolute rule is that the first constant " $a$ " cannot be a zero.
In the given equation we have the variable $p$ in place of $x$ . And the numerical coefficients $a$ , $b$ , and $c$ are replaced by $4$ , $7$ and $3$ respectively.
Hence, we can say that $a=4$ , $b=7$ and $c=3$
Now, we use the Sridhar Acharya’s formula or the quadratic formula which is
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Replacing all the terms we have mentioned previously we get
\[\Rightarrow p=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times 4\times 3}}{2\times 4}\]
\[\Rightarrow p=\dfrac{-7\pm \sqrt{1}}{8}\]
Taking both the signs into account we get
\[\Rightarrow p=\dfrac{-7+1}{8}\]
\[\Rightarrow p=-\dfrac{6}{8}\]
\[\Rightarrow p=-\dfrac{3}{4}\]
And
 \[\Rightarrow p=\dfrac{-7-1}{8}\]
\[\Rightarrow p=\dfrac{-8}{8}\]
\[\Rightarrow p=-1\]

Therefore, the solution of the equation is \[p=-1\] and \[p=-\dfrac{3}{4}\]

Note: While solving these types of equations we must keep in mind that we have to take both the positive and negative signs into account to get all the solutions of the equation. Also, we must properly put the values of $a$ , $b$ , and $c$ in the formula to get the proper solutions.