Solve the given equation ${x^2} - 2x - 9 = 0$.

Hint: In this equation, we will use the rule of Sridhar Acharya that is if the quadratic equation will be in the form $a{x^2} + bx + c = 0$ then $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . For this question at first we will compare the coefficients of ${x^2}$ , $x$ and the constant part with $a{x^2} + bx + c = 0$ and then we will solve the equation by the Sridhar Acharya rule.

Complete step-by-step solution:
We have;
${x^2} - 2x - 9 = 0$
Comparing this equation with $a{x^2} + bx + c = 0$ we will get;
$a = 1$
And $b = - 2$
And $c = - 9$ .
Now we will apply the Sridhar Acharya method and put these values in the equation $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
We will get;
$x = \dfrac{{ - ( - 2) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \cdot 1 \cdot ( - 9)} }}{{2 \cdot 1}}$
Simplifying the above equation we will get;
$\Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 4 \cdot 1 \cdot 9} }}{{2 \cdot 1}}$
Simplifying the above equation we will get;
$\Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 36} }}{2}$
After adding we get;
$\Rightarrow x = \dfrac{{2 \pm \sqrt {40} }}{2}$
Now we will factorize $\;40$ and found out all the numbers that are twice. Then those terms will come out of the square root sign.
The factors of $\;40$ are $2$ , $2$ , $2$ , $5$ .
So we can write;
$\Rightarrow x = \dfrac{{2 \pm \sqrt {2 \cdot 2 \cdot 2 \cdot 5} }}{2}$
Simplifying the above equation we get;
$\Rightarrow x = \dfrac{{2 \pm 2\sqrt {2 \cdot 5} }}{2}$
After multiplying we get;
$\Rightarrow x = \dfrac{{2 \pm 2\sqrt {10} }}{2}$
Dividing the numerator with the denominator we will get;
$\Rightarrow x = 1 \pm \sqrt {10}$

So our solution is $x = 1 + \sqrt {10}$
and $x = 1 - \sqrt {10}$.

Additional Information: We have to always remember that in the quadratic equation we will always get two values of the variable. It may happen we will get a single value of the variable. That means the quadratic equation has a double root. So we have to write in the answer twice. The quadratic equation can give us the complex root whenever there will be a negative value under the square root.

Note: This kind of equation with the square of $x$ is called a quadratic equation. In the Sridhar Acharya method, the ‘ $\pm$ ‘ implies two values of $x$ are expected from the equation. If there are any numbers under the square root sign which are not a perfect square of any integer we will leave the number under the square root like this question.