How do you solve $3{{x}^{2}}+4x-3=0$ using the quadratic formula?

Hint: As the given is the quadratic equation and we have to solve it by using the quadratic formula. So we know that for a equation of the form $a{{x}^{2}}+bx+c=0$ the quadratic formula is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . by substituting the values and solving further we get the desired answer.

Complete step-by-step solution:
We have been given an equation $3{{x}^{2}}+4x-3=0$.
We have to solve the given equation by using the quadratic formula.
The name quad means square as the variable has power two so the given equation is quadratic equation. We know that quadratic formula for the general equation $a{{x}^{2}}+bx+c=0$ is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Now, by comparing the given equation with general equation we get the values
$a=3,b=4,c=-3$
Now, substituting the values in the formula we will get
$\Rightarrow x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 3\times \left( -3 \right)}}{2\times 3}$
Now, simplifying the above equation we will get
$\begin{align}
  & \Rightarrow x=\dfrac{-4\pm \sqrt{16+36}}{6} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{52}}{6} \\
\end{align}$
Now, simplifying the above obtained equation we will get
$\Rightarrow x=\dfrac{-4\pm 4\sqrt{13}}{6}$
Taking the common terms out and simplifying further we will get
$\Rightarrow x=\dfrac{-2\pm 2\sqrt{13}}{3}$
Hence we get the solution of given quadratic equation as $x=\dfrac{-2+2\sqrt{13}}{3},\dfrac{-2-2\sqrt{13}}{3}$.

Note: The value of x is the solution of the quadratic equation also known as the roots of the equation. The roots of the quadratic equation can be real numbers or complex numbers. As the degree of the given equation is two so the equation has two solutions i.e. two values of x. Also if the method is not mentioned in the question we can solve the quadratic equation by using completing the square method or by split middle term method.