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

- Hint: To solve this type of problem first we have to find the value of discriminant and say how the roots will be. Then we have to write the values in the formula to get the roots. Nature of the roots is given by \[{{b}^{2}}-4ac\]. By substituting the values of a, b, c in this formula gives us the roots. \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step-by-step solution -

Given equation is \[3{{x}^{2}}-7x+2=0\].
The value of a, b, c is noted.
\[a=3,b=-7,c=2\]
Discriminant: \[{{b}^{2}}-4ac\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Substituting the values in (1) we get,
\[{{\left( -7 \right)}^{2}}-4\left( 3 \right)\left( 2 \right)>0\]
That means the roots are real and unequal.

To find the roots we have the formula,
\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now substituting the values of a, b, c in (2) we get,
$\Rightarrow$ \[\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\left( 3 \right)\left( 2 \right)}}{2\left( 3 \right)}\]
$\Rightarrow$ \[\dfrac{7\pm \sqrt{25}}{6}\]
$\Rightarrow$ \[\dfrac{7+\sqrt{25}}{6},\dfrac{7-\sqrt{25}}{6}\]
$\Rightarrow$ \[\dfrac{7+5}{6},\dfrac{7-5}{6}\]
$\Rightarrow$ \[\dfrac{12}{6},\dfrac{2}{6}\]

The roots for the quadratic equation is \[2,\dfrac{1}{3}\].

Note: In (2) the term under root is not zero. The discriminant plays a major role which specifies the nature of roots. The roots can be either real or imaginary. This is a direct problem with using mathematical operations. Be careful while doing calculations.