Question 856303
First, you set up an equation and solve it.
Then, you discard any solution that does not fit the description.
(In this case the description is "integer",
but some problems refer to "even integer", or "odd integer").
 
You need to define your "two consecutive integers".
You may write {{{n}}}= the first integer,
to define your variable.
Or you could also use {{{x}}} for your variable.
 
Using {{{x}}}, you know (and may, or may not want to write) that:
{{{x+1}}}= the second integer
{{{x^2}}}= the square of the first integer,
{{{3x^2}}}= three times the square of the first integer,
{{{3x^2+3}}}= three times the square of the first integer plus three, and
{{{5(x+1)}}}= 5 times the second integer
 
Your equation is
{{{3x^2+3=5(x+1)}}}
 
Then you work on rearranging and simplifying your equation as needed:
{{{3x^2+3=5(x+1)}}}
{{{3x^2+3=5x+5}}}
{{{3x^2+3-5x-5=0}}}
{{{3x^2-5x-2=0}}}
 
Finally, you solve by a method of your choice.
In this case, I would use the quadratic formula,
which says that the solutions to
{{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}.
In this case, applying the quadratic formula, we get
{{{x = (-(-5) +- sqrt((-5)^2-4*3*(-2) ))/(2*3) }}}
{{{x = (5 +- sqrt(25+24 ))/6}}}
{{{x = (5 +- sqrt(49))/6}}}
{{{n = (5 +- 7)/6}}}
You get the solutions
{{{x = (5 +7)/6}}}-->{{{x=12/6}}}-->{{{highlight(x=2)}}}
and {{{x = (5 -7)/6}}}-->{{{x=-2/6}}}-->{{{x=-1/3}}}
Since {{{-1/3}}} is not an integer, it is not a solution to the problem, and you discard it.
The two consecutive integers are {{{x=highlight(2)}}} and {{{x+1=2+1=highlight(3)}}} .
 
Verification:
{{{3*2^2+3=3*4+3=12+3=15}}} would be three times the square of the first integer plus three;
{{{5*3=15}}} would be "5 times the second integer",
and both quantities are equal, as the problem required.