How do you find two consecutive integers whose product is 783?

Hint: In this question, we have to find the two consecutive numbers whose product is equal to 783. Two numbers that are next to each other, that is, one number is one more than the other, or we can also suppose one number to be smaller than the other. By letting one number be “x”, the other number can be supposed to be “x+1” or “x-1”. Then the product of these two numbers is given as 783, so we can form an equation and solve the obtained equation using the appropriate method.

Complete step-by-step answer:
We are given that
\[
  x(x + 1) = 783 \\
   \Rightarrow {x^2} + x = 783 \\
   \Rightarrow {x^2} + x - 783 = 0 \;
\]
Now the obtained equation is quadratic, so we can solve it by quadratic formula as follows –
 $
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
  x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)( - 783)} }}{{2(1)}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 3132} }}{2} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {3133} }}{2} \\
   \Rightarrow x \cong \dfrac{{ - 1 \pm 55.973}}{2} \\
   \Rightarrow x \cong \dfrac{{ - 56.973}}{2},\,x \cong \dfrac{{54.973}}{2} \\
   \Rightarrow x \cong - 28.487,\,x \cong 27.487 \;
 $
These are the two numbers whose product is equal to 783, but these numbers are not integers so we can say that there are no two consecutive integers whose product is 783.
However, there are two consecutive “odd” integers whose product is equal to 783 –
 $ 27 \times 29 = 783 $

Note: Integers are a type of real numbers, they can be positive as well as negative, and these numbers do not involve decimal point or fraction, that’s why the obtained solutions of the equation are rejected. A quadratic equation can be solved by using factorization/quadratic formula/completing the square method/graph. When the equation cannot be factorized, the equation is solved by using the quadratic formula as in this question.