How do you solve $2{{x}^{2}}-7x+3=0$ ?

Hint: The equation given in the question is a quadratic equation , we know the roots of a quadratic equation can calculated by using quadratic formula which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term in a quadratic equation.

Complete step by step answer:
The given equation in the question is $2{{x}^{2}}-7x+3=0$ which is quadratic equation
If we compare the equation $2{{x}^{2}}-7x+3=0$ to $a{{x}^{2}}+bx+c=0$ the value of a is 2, the value of b is -7 and the value of c is equal to 3
We know the formula for roots of a quadratic equation which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We put the value of a, b, and c in the above formula we can get the value of roots
So the roots are $\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 2\times 3}}{2\times 2}$
Further solving we get
$\Rightarrow \dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 2\times 3}}{2\times 2}=\dfrac{7\pm 5}{4}$
So the roots are 3 and $\dfrac{1}{2}$

Note:
We can solve the quadratic equation in factorization method. In this method we can write the quadratic equation $a{{x}^{2}}+bx+c=0$ as $a\left( x-\alpha \right)\left( x-\beta \right)=0$ where $\alpha $ and $\beta $ are roots of the equation. Let’s solve $2{{x}^{2}}-7x+3=0$ by method of factorization.
We can write $2{{x}^{2}}-7x+3=0$ = $2{{x}^{2}}-6x-x+3=0$
We can take 2x common in the first half and -1 common in the second half of the equation
$\Rightarrow 2{{x}^{2}}-7x+3=2x\left( x-3 \right)-1\left( x-3 \right)$
We can take x- 3 common from the whole equation
$\Rightarrow 2{{x}^{2}}-7x+3=\left( 2x-1 \right)\left( x-3 \right)$
If $\left( 2x-1 \right)\left( x-3 \right)$ is equal to 0 then x can be 3 or $\dfrac{1}{2}$
So the roots of the equation $2{{x}^{2}}-7x+3=0$ are 3 and $\dfrac{1}{2}$