How do you solve $20{x^2} - 41x + 20$?

Hint:
In this question we have to factorise the given polynomial using quadratic formula. In the polynomial $a{x^2} + bx + c$, where "$a$", "$b$", and “$c$" are real numbers and the Quadratic Formula is derived from the process of completing the square, and is formally stated as:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
Quadratic equations are the equations that are often called second degree. It means that it consists at least one term which is squared, the general form of quadratic equation is $a{x^2} + bx + c = 0$, where "$a$", "$b$", and "$c$" are numerical coefficients or constant, and the value of $x$ is unknown. And one fundamental rule is that the value of $a$, the first constant cannot be zero in a quadratic equation.
Now the given quadratic equation is,
$20{x^2} - 41x + 20$,
Now using the quadratic formula, which is given by$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
Here$a = 20$,$b = - 41$,$c = 20$,
Now substituting the values in the formula we get,
$ \Rightarrow x = \dfrac{{ - \left( { - 41} \right) \pm \sqrt {{{\left( { - 41} \right)}^2} - 4\left( {20} \right)\left( {20} \right)} }}{{2\left( {20} \right)}}$,
Now simplifying we get,
$ \Rightarrow x = \dfrac{{41 \pm \sqrt {1681 - \left( {1600} \right)} }}{{40}}$,
Now again simplifying we get,
$ \Rightarrow x = \dfrac{{41 \pm \sqrt {1681 - 1600} }}{{40}}$,
Further simplification we get,
$ \Rightarrow x = \dfrac{{41 \pm \sqrt {81} }}{{40}}$,
Now taking the square root we get,
$ \Rightarrow x = \dfrac{{41 \pm 9}}{{40}}$,
Now we get two values of $x$ they are $x = \dfrac{{41 + 9}}{{40}} = \dfrac{{50}}{{40}} = \dfrac{5}{4}$and
,$x = \dfrac{{41 - 9}}{{40}} = \dfrac{{32}}{{40}} = \dfrac{4}{5}$.
So the values of $x$are $\dfrac{5}{4}$ and $\dfrac{4}{5}$.

$\therefore $If we solve the given equation, i.e., $20{x^2} - 41x + 20$, then the values of $x$ are $\dfrac{5}{4}$ and $\dfrac{4}{5}$.

Note:
Quadratic equation formula is a method of solving quadratic equations, but we should keep in mind that we can also solve the equation using completely the square, and we can cross check the values of $x$ by using the above formula. Also we should always convert the coefficient of ${x^2} = 1$, to easily solve the equation by this method, and there are other methods to solve such kind of solutions, other method used to solve the quadratic equation is by factoring method, in this method we should obtain the solution factorising quadratic equation terms. In these type of questions, we can solve by using quadratic formula i.e.,$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.