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**Hint:**By quadratic formula, we mean the Sridhar Acharya formula, which is more commonly known. We compare the given equation with the general form of a quadratic equation $a{{x}^{2}}+bx+c=0$ to find out the coefficients and put them in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . After that, we do some basic arithmetic operations and get two solutions to the given equation.

**Complete step by step solution:**

The given equation is

$3{{x}^{2}}-12x+11=0$

Now, let us consider the general form of a quadratic equation which is

$a{{x}^{2}}+bx+c=0$

By quadratic formula, we mean the Sridhar Acharya formula which states that the two roots of a quadratic equation can be expressed as

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Where, $a,b$ are the respective coefficients of the terms ${{x}^{2}}$ , $x$ and $c$ is the constant term. Comparing the given equation $3{{x}^{2}}-12x+11=0$ with the general quadratic equation $a{{x}^{2}}+bx+c=0$ , we get

$\begin{align}

& a=3 \\

& b=-12 \\

& c=11 \\

\end{align}$

Having found out the coefficients, we put them in the Sridhar Acharya formula to get the two roots of the equation or the two values of $x$ that satisfy the given equation. Thus,

$\Rightarrow x=\dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\left( 3 \right)\left( 11 \right)}}{2\left( 3 \right)}$

Simplifying the terms, we get,

\[\Rightarrow x=\dfrac{12\pm \sqrt{144-4\left( 33 \right)}}{6}\]

Further simplification gives

$\Rightarrow x=\dfrac{12\pm \sqrt{144-132}}{6}$

Subtracting the two numbers $144$ and $132$ and implementing the square root, we get,

$\Rightarrow x=\dfrac{12\pm 2\sqrt{3}}{6}$

Now, the two values of $x$ after separating the $+$ and the $-$ from the $\pm $ , are

$\Rightarrow x=\left( \dfrac{12+2\sqrt{3}}{6} \right)$ and $x=\left( \dfrac{12-2\sqrt{3}}{6} \right)$

On simplification, we get,

$\Rightarrow x=2.577$ and $x=1.422$

**Therefore, we can conclude that the two roots or solutions of the given equation $3{{x}^{2}}-12x+11=0$ , found out by the quadratic formula are $x=2.577$ and $x=1.422$**

**Note:**We should remember the quadratic formula accurately and should not interpret it wrongly. As the coefficients play a vital role here, we must compare our equation with the general quadratic equation correctly and any mistake in involving the signs will lead to wrong answers. Also, there are various other ways to find the solution like, by plotting the corresponding graph and see where it intersects the $x-axis$ , by completing the square, or by the middle term method, or the vanishing factor method.

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