How do you factor and solve $ {{x}^{2}}+4x-3=0 $ ?

Hint: To factorize the given quadratic equation, we are going to use the Shree Dharacharya rule. From Shree Dharacharya rule for the quadratic equation $ a{{x}^{2}}+bx+c=0 $ we can find the roots of the quadratic equation as follows: $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . Now, we will get two values of x (positive and negative). And further, we will get the factors for the given quadratic equation.

Complete step by step answer:
The quadratic equation given in the above problem of which we have to find the factors is equal to:
 $ {{x}^{2}}+4x-3=0 $
Applying Shree Dharacharya rule in the above quadratic equation we get,
 $ x=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( -3 \right)}}{2\left( 1 \right)} $
Taking 4 as common from the square root we get,
 $ \begin{align}
  & x=\dfrac{-4\pm \sqrt{4\left( 4-\left( 1 \right)\left( -3 \right) \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-4\pm 2\sqrt{\left( 4+3 \right)}}{2\left( 1 \right)} \\
\end{align} $
Now, 2 will be cancelled out from the numerator and the denominator and we get,
 $ x=\dfrac{-2\pm \sqrt{7}}{1} $
Taking the value of x with positive sign we get,
 $ \begin{align}
  & x=-2+\sqrt{7} \\
 & \Rightarrow x+2-\sqrt{7}=0 \\
\end{align} $
Taking the value of x with negative sign we get,
 $ \begin{align}
  & x=-2-\sqrt{7} \\
 & \Rightarrow x+2+\sqrt{7}=0 \\
\end{align} $
Now, we can write the roots of quadratic equation in the factors we get,
 $ \left( x+2-\sqrt{7} \right)\left( x+2+\sqrt{7} \right) $
Hence, we have factorized the given quadratic equation in the following way:
 $ \left( x+2-\sqrt{7} \right)\left( x+2+\sqrt{7} \right) $

Note:
 You might be thinking that we can find the factors of giving equation $ {{x}^{2}}+4x-3=0 $ in the following way:
By taking the prime factorization of 3 and then add and subtract the factors of 3 in such a way so that we will get 4 and then write the addition or subtraction of factors of 3 in place of 4. After that, we will simplify the expression just obtained to get the factors for the given quadratic equation. But while factorizing in this manner, you will encounter the following problem:
Now, to factorize the above quadratic equation, first of all, we are going to take the factorization of 3.
Number 3 has only two factors i.e. 1 and the number itself.
 $ 3=3\times 1 $
After that, if we add the factors 3 and 1 we will get 4 so replacing 4 with $ (3+1) $ we get,
 $ {{x}^{2}}+\left( 3+1 \right)x-3=0 $
Opening the bracket in the above equation we get,
 $ {{x}^{2}}+3x+x-3=0 $
Now, taking x as common from the first two terms we get,
 $ x\left( x+3 \right)+1\left( x-3 \right)=0 $
Here, the problem will come because there is no common expression in the above equation so this method has failed.