How do you solve \[ - 2{x^2} = 2x - 4\] ?

Hint: In this question, a polynomial equation of degree 2 is given to us as 2 is the highest exponent in the given equation, hence it is a quadratic equation and has exactly two solutions. Solutions of an equation are defined as the values of the x for which the given function has a value zero or when we plot this function on the graph, we see that the solutions of this function are the points on which the y-coordinate is zero, thus they are simply the x-intercepts.

Complete step-by-step solution:
We are given that
\[ - 2{x^2} = 2x - 4\]
Now we will take all the terms to one side such that they are equal to zero and then take 2 and negative sign common to convert it into standard form –
$
   - 2{x^2} - 2x + 4 = 0 \\
   \Rightarrow {x^2} + x - 2 = 0 \\
 $
The obtained equation is in standard form, on factorization we get –
$
  {x^2} + x - 2 = 0 \\
   \Rightarrow {x^2} + 2x - x - 2 = 0 \\
   \Rightarrow x(x + 2) - 1(x + 2) = 0 \\
   \Rightarrow (x - 1)(x + 2) = 0 \\
   \Rightarrow x - 1 = 0,\,x + 2 = 0 \\
   \Rightarrow x = 1,\,x = - 2 \\
 $
Hence the factors of the equation \[ - 2{x^2} = 2x - 4\] are $x - 1 = 0$ and $x + 2 = 0$ .

Note: The standard form of a quadratic equation is $a{x^2} + bx + c = 0$ . To find the factors of the given equation, we compare the given equation and the standard equation and get the values of a, b and c. Then we will try to write b as a sum of two numbers such that their product is equal to the product of a and c, that is, ${b_1} \times {b_2} = a \times c$ , this method is known as factorization. We find the value of ${b_1}$ and ${b_2}$ by hit and trial. We move to some other methods like quadratic formula, graphing, and completing the square method, if we are not able to solve an equation by factorization.