How do you factor \[{x^2} + 5x - 6?\]

Hint: The given question describes the operation of using the quadratic formula, addition/ subtraction/ multiplication/ division. Also, remind the quadratic formula and compare the given equation with the quadratic formula to solve the given equation. We need to know the basic form of the quadratic equation to find the value \[x\] .

Complete step-by-step answer:
The given question is shown below,
 \[{x^2} + 5x - 6 = 0 \to \left( 1 \right)\]
We know that the basic form of the quadratic formula is,
 \[a{x^2} + bx + c = 0 \to \left( 2 \right)\]
Then,
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
Let’s compare the equation \[\left( 1 \right)and\left( 2 \right)\] for finding the value of \[a,b,\] and \[c\]
 \[
  \left( 1 \right) \to {x^2} + 5x - 6 = 0 \\
  \left( 2 \right) \to a{x^2} + bx + c = 0 \;
\]
Let’s compare the \[{x^2}\] terms in the equation \[\left( 1 \right)and\left( 2 \right)\] we get,
 \[
  1 \times {x^2} \\
  a \times {x^2} \;
\]
So, the value of \[a = 1\]
Let’s compare the \[x\] terms in the equations \[\left( 1 \right)and\left( 2 \right)\]
 \[
  5 \times x \\
  b \times x \;
\]
So, the value of \[b\] \[ = 5\]
Let’s compare the constant terms in the equation \[\left( 1 \right)and\left( 2 \right)\]
 \[
   - 6 \\
  c \;
\]
So, the value of \[c = - 6\]
So, we get \[a,b\] and \[c\] values are \[1,5\] and \[ - 6\] respectively. Let’s substitute these values in the equation \[\left( 3 \right)\] for finding the values of \[x\]
 \[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
By substituting the value of \[a,b\] and \[c\] in the equation \[\left( 3 \right)\] we get,
 \[x = \dfrac{{ - \left( 5 \right) \pm \sqrt {\left( {{5^2}} \right) - 4\left( 1 \right)\left( { - 6} \right)} }}{{2 \times 1}}\]
By solving the above equation, we get
 \[x = \dfrac{{ - \left( 5 \right) \pm \sqrt {25 + 24} }}{2}\]
Let’s add the two terms inside the root, we get
 \[x = \dfrac{{ - \left( 5 \right) \pm \sqrt {49} }}{2}\]
We know that \[\sqrt {49} \] can also be written as \[{7^2}\] .
The square and square roots are cancelled each other, so we get
 \[x = \dfrac{{\left( { - 5 \pm 7} \right)}}{2}\]
Due to the \[ \pm \] operation, we get two types of values for \[x\]
Case: 1
 \[x = \dfrac{{ - 5 + 7}}{2}\]
 \[x = \dfrac{2}{2}\]
 \[x = 1\]
Case: 2
 \[x = \dfrac{{ - 5 - 7}}{2}\]
 \[
  x = \dfrac{{ - 12}}{2} \\
  x = - 6 \;
\]
In case: 1 we assume \[ \pm \] as \[ + \] and in case: 2 we assume \[ \pm \] as \[ - \] . By considering the \[ \pm \] as \[ + \] we get the value of \[x\] is \[1\] . By considering the \[ \pm \] as \[ - \] we get the value of \[x\] is \[ - 6\] .
So, the final answer is
 \[
  x = 1(or) \\
  x = - 6 \;
\]
So, the correct answer is x=1 OR x=-6”.

Note: To find the value \[x\] from the given equation we would compare the equation with the basic form of a quadratic equation. After comparing the equation we would find the value of \[a,b\] , and \[c\] . When substituting these values in the quadratic formula remind the following things,
I.When a negative number is multiplied by another negative number, the final answer becomes positive.
II.When a negative number is multiplied by a positive number, the final answer becomes negative.
III.When a positive number is multiplied by a positive number, the final answer becomes positive.