How do you solve it\[{x^2} + 2x - 120 = 0\]?

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 a quadratic formula to solve the given equation. Also, take care of the multiplication part.

Complete step by step answer:
The given equation is shown below,
\[{x^2} + 2x - 120 = 0 \to \left( 1 \right)\]
We know that 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 in the given equation,
\[\left( 1 \right) \to {x^2} + 2x - 120 = 0\]
\[\left( 2 \right) \to a{x^2} + bx + c = 0\]
So, let’s compare the \[{x^2}\] terms in the equation \[\left( 1 \right)and\left( 2 \right)\]
\[
  1 \times {x^2} \\
  a \times {x^2} \\
    \\
 \]
So, we find the value of a which is,
\[a = 1\]
Let’s compare the \[x\] terms in the equations \[\left( 1 \right)and\left( 2 \right)\]
\[
  2 \times x \\
  b \times x \\
 \]
So, we find the value of b, which is
\[b = 2\]
Let’s compare the constant terms in the equation \[\left( 1 \right)and\left( 2 \right)\]
\[
   - 120 \\
  c \\
 \]
So, we find the value of b, which is
\[c = - 120\]
So, we get a, b, and c values are 1, 2, and -120 respectively. Let’s substitute these values in the equation \[\left( 3 \right)\] for finding the value of x
\[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
By substituting the values of a, b, and c in the equation \[\left( 3 \right)\]the above equation become,
\[x = \dfrac{{ - \left( 2 \right) \pm \sqrt {{{\left( 2 \right)}^2} - 4 \times 1 \times \left( { - 120} \right)} }}{{2 \times 1}}\]
Let’s solve the above equation, we get
\[x = \dfrac{{ - \left( 2 \right) \pm \sqrt {4 + 480} }}{2}\]
Let’s add the two terms inside the root, we get
\[x = \dfrac{{ - 2 \pm \sqrt {484} }}{2}\]
We know that \[\sqrt {484} \] can also be written as 22 because\[\left( {{{22}^2} = 484} \right)\]- So, the square and square root are canceled each other.
\[x = \dfrac{{ - 2 \pm 22}}{2}\]
Due to the” \[ \pm \] “we get two values for x
Case (1):
\[
    \\
  x = \dfrac{{20}}{2} \\
  x = 10 \\
 \]
\[
    \\
  x = \dfrac{{20}}{2} \\
  x = 10 \\
 \]
 Case (2):
  \[
  x = \dfrac{{ - 2 - 22}}{2} \\
  x = \dfrac{{ - 24}}{2} \\
  x = - 12 \\
    \\
 \]
So, the final answer is \[x = 10or - 12\].

Note:
To find the value of x from the given equation we would compare the equation with the quadratic formula. After comparing the equation we would find the value of a, b, and c. 1) When substituting these values in the quadratic formula remind the following things,
1) When a negative number multiplied with the negative number becomes the positive number.
2) When a positive number multiplied with another positive number the answer becomes positive.
3) Anyone's term is negative, the answer becomes negative.
The square root of 1 becomes 1