Solve the quadratic equation
 \[3{x^2} - 7x + 2 = 0\]

Hint: If we have a polynomial of degree ‘n’ then we have ‘n’ number of roots or factors. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula, etc. The quadratic formula is used when we fail to find the factors of the equation. We are going to solve this using the quadratic formula method. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[3{x^2} - 7x + 2 = 0\].
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We get, \[a = 3\], \[b = - 7\] and \[c = 2\].
Substituting in the formula of standard quadratic, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
\[ \Rightarrow x = \dfrac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(3)(2)} }}{{2(3)}}\]
\[ \Rightarrow x = \dfrac{{7 \pm \sqrt {49 - 24} }}{6}\]
\[ \Rightarrow x = \dfrac{{7 \pm \sqrt {25} }}{6}\]
We know that 25 is a perfect square.
\[ \Rightarrow x = \dfrac{{7 \pm 5}}{6}\]
Thus, we have two roots,
\[ \Rightarrow x = \dfrac{{7 + 5}}{6}\] and \[x = \dfrac{{7 - 5}}{6}\]
\[ \Rightarrow x = \dfrac{{12}}{6}\] and \[x = \dfrac{2}{6}\]
\[ \Rightarrow x = 2\] and \[x = \dfrac{1}{3}\]
This is the required result.

Note: We can also solve this using the factorization method. We have \[3{x^2} - 7x + 2 = 0\]
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that\[{b_1} \times {b_2} = ac\] and\[{b_1} + {b_2} = b\].
Here we can say that \[{b_1} = - 6\] and \[{b_2} = - 1\]. Because \[{b_1} \times {b_2} = 6\] \[(a \times c)\] and \[{b_1} + {b_2} = - 7(b)\].
Now we write \[3{x^2} - 7x + 2 = 0\] as,
\[3{x^2} - 6x - x + 2 = 0\]
Taking ‘3x’ common in the first two terms and taking $-1$ common in the remaining two terms we have,
\[ \Rightarrow 3x(x - 2) - (x - 2) = 0\]
Again taking \[(x - 2)\] common we have,
\[ \Rightarrow (3x - 1)(x - 2) = 0\].
By zero product principle we have,
 \[ \Rightarrow (3x - 1) = 0\] and \[(x - 2) = 0\]
\[ \Rightarrow 3x = 1\] and \[x = 2\]
\[ \Rightarrow x = \dfrac{1}{3}\] and \[x = 2\].
This is the required solution. We can see that in both we have the same answer.