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**Hint:**We have been given a quadratic equation of $a$ as $3{{a}^{2}}=6a-3$. We use the quadratic formula to solve the value of the $a$. we have the solution in the form of $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$ for general equation of $p{{x}^{2}}+qx+r=0$. We put the values and find the solution.

**Complete step-by-step solution:**

We know for a general equation of quadratic $p{{x}^{2}}+qx+r=0$, the value of the roots of $x$ will be $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$. This is the quadratic equation solving method. The root part $\sqrt{{{q}^{2}}-4pr}$ of $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$ is called the discriminant of the equation.

In the given equation we have $3{{a}^{2}}-6a+3=0$. The values of p, q, r are $3,-6,3$ respectively.

We put the values and get $a$ as \[a=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 3\times 3}}{2\times 3}=\dfrac{6\pm 0}{6}=\dfrac{6}{6}=1\]

**The roots of the equation are real numbers. Two roots being equal in value as the expression is a square form. So, values of $a$ are $a=1$.**

**Note:**We can also apply the middle-term factoring or grouping to factorise the polynomial.

We have

$\begin{align}

& 3{{a}^{2}}=6a-3 \\

& \Rightarrow {{a}^{2}}-2a+1=0 \\

& \Rightarrow {{\left( a-1 \right)}^{2}}=0 \\

\end{align}$

Now we take square root both sides of the equation and get

$\begin{align}

& {{\left( a-1 \right)}^{2}}=0 \\

& \Rightarrow a-1=0 \\

\end{align}$

The root of the equation becomes 1 where two roots being equal in value.

Therefore, values of $a$ are $a=1$.

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