How do you factor $ {x^2} + 4x - 5 $ ?

Hint:Here we will use the standard quadratic equation and will compare the given equation with it and then will find the roots of the given equation using the formula \[x = \dfrac{{ - b
\pm \sqrt \Delta }}{{2a}}\] .

Complete step by step solution:
Again, take the given expression –
 $ {x^2} + 4x - 5 $
Let us consider the general form of the quadratic equation.
 $ a{x^2} + bx + c = 0 $
Compare the standard equation with the given equation $ {x^2} + 4x - 5 $ .....(A)
 $
\Rightarrow a = 1 \\
\Rightarrow b = 4 \\
\Rightarrow c = - 5 \\
 $
Now, $ \Delta = {b^2} - 4ac $
Place the values in the above equation –
 $ \Delta = {4^2} - 4(1)( - 5) $
Simplify the above equation –
 $ \Delta = 16 + 20 $
Simplify the above equation –
 $ \Delta = 36 $
Take root on both the sides of the equation –
 $ \Rightarrow \sqrt \Delta = \sqrt {36} $
 $ \Rightarrow \sqrt \Delta = 6 $
Now, the roots can be expressed as
 \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]

Therefore, \[x = \dfrac{{ - b + \sqrt \Delta }}{{2a}}\] or \[x = \dfrac{{ - b - \sqrt \Delta
}}{{2a}}\]
Place the known values in the above equation –
 \[x = \dfrac{{ - (4) + 6}}{2}\] or \[x = \dfrac{{ - (4) - 6}}{2}\]
When you add one positive number and negative number you have to do subtraction and sign of a bigger number. And when you have to subtract from a negative term, you have to add and apply signs of negative.
 \[ \Rightarrow x = \dfrac{2}{2}\] or \[ \Rightarrow x = \dfrac{{ - 10}}{2}\]
Simplify the above equation –
 \[ \Rightarrow x = \dfrac{{10}}{2}\] or \[ \Rightarrow x = \dfrac{{ - 2}}{2}\]
Common multiple from the numerator and the denominator cancel each other.
 \[ \Rightarrow x = 5\] or \[ \Rightarrow x = ( - 1)\]
This is the required solution.

Note: The above example can be solved by another method.
Take the given expression –
 $ {x^2} + 4x - 5 $
Here we have three terms in the given expression.
Now, multiply the constant in the first term with the last term.
i.e. $ 1 \times ( - 5) = ( - 5) $
Now, you have to split the middle term to get $ ( - 5) $ in multiplication and addition or subtraction to get middle term i.e. $ 4 $ .
 $
( - 5) = (5) \times ( - 1) \\
4 = (5) - 1 \\
 $ $ $ $ $
Write the equivalent value for the middle term
 $ {x^2} + \underline {5x - x} - 5 $
 $ = \underline {{x^2} + 5x} + \underline {x - 5} $
Take common factors from the above paired terms.
 $ = x(x + 5) - 1(x + 5) $
Take common factors-
 $ = (x + 5)(x - 1) $
The above expression –
 $ \Rightarrow x + 5 = 0 $ or $ \Rightarrow x - 1 = 0 $
The above expression applies –
 $ \Rightarrow x = ( - 5) $ or $ \Rightarrow x = 1 $