Solve for $x$ : $10{x^2} - 27x + 5 = 0$

Hint: Given equation is in linear form with one variable $x$ and the highest power of $x$ is 2. These kinds of equations are called quadratic equations. These equations can be solved directly by converting them into standard form and using a formula or elaborating the equation further to convert the equation into a multiple of two different equations each of a single degree and equating them individually to zero to get two different values for $x$ .
Formula used: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete answer:
Step 1:
The given equation is a quadratic equation. The values of $x$ at which the given equation is satisfied are called roots of the equation.
A quadratic equation may have either two real roots or two imaginary roots.
Any quadratic equation can be rearranged in the form of $a{x^2} + bx + c = 0$
For a quadratic equation of this form, the roots are given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Step 2:
The given equation is $10{x^2} - 27x + 5 = 0$ which is similar to the standard form of $a{x^2} + bx + c = 0$
Thus, on comparing the given equation and the standard form, we get
$a = 10$
$b = - 27$
$c = 5$
Step 3:
The roots of the above equation can be obtained by substituting the values of $a$ , $b$ and $c$ in the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Upon substitution, we get
$x = \dfrac{{ - \left( { - 27} \right) \pm \sqrt {{{\left( { - 27} \right)}^2} - 4 \times 10 \times 5} }}{{2 \times 10}}$
$x = \dfrac{{27 \pm \sqrt {729 - 200} }}{{20}}$
$x = \dfrac{{27 \pm \sqrt {529} }}{{20}} = \dfrac{{27 \pm 23}}{{20}}$
To get the solutions, we divide the based upon + and – symbols
So, we get
${x_1} = \dfrac{{27 + 23}}{{20}} = \dfrac{{50}}{{20}} = \dfrac{5}{2}$ and
${x_2} = \dfrac{{27 - 23}}{{20}} = \dfrac{4}{{20}} = \dfrac{1}{5}$
Thus we get 2 solutions each of different values as ${x_1} = \dfrac{5}{2}$ and ${x_2} = \dfrac{1}{5}$ .

Note:
In the above problem, we got two different roots for $x$ . But in the case of some other quadratic equations, we don’t get two unique solutions. This can be found out through the value of ${b^2} - 4ac$ . If the value of ${b^2} - 4ac$ is greater than or equal to $0$ , we get two real roots and if it’s value is less than $0$ , we get two imaginary roots.