Can you tell that for example it's an equation #y=2x^2-6x+5# And question is to find the set values of x for which y>13 So in it answers are x<-1 and x>4 How do we know that x is greater or less than -1 or 4 as when we solve it on calculator?

We have #y=2x^2-6x+5# and we seek values of #x# for which #y>13#

#y>13# means #2x^2-6x+5>13#

i.e. #2x^2-6x+5-13>0#

or #2x^2-6x-8>0#

or #x^2-3x-4>0#

or #(x-4)(x+1)>0#

This means product of #(x-4)# and #(x+1)# is positive.

One when both are positive i.e. #x-4>0# and #x+1>0# and in other words #x>4# and #x>-1#. This is possible only when #x>4#

Other possibility is when both are negative i.e. #x-4<0# and #x+1<0# and in other words #x<4# and #x<-1#. This is possible only when #x<-1#

Hence #y>13#, when either #x<-1# or when #x>4#

In fact if we draw graph of #y=2x^2-6x+5#, we see that #y>13# when either #x<-1# or when #x>4#

set #2x^2-6x+5=y>13#

As the #2x^2# is positive then the graph is of general shape #uu#

By changing the above from > to = we can solve for the points where #y# is just starting to be #y>13#

To find the critical points write #2x^2-6x+5=13# as:

#2x^2-6x+5-13=0#

The required #x# values will be where #x# satisfies #2x^2-6x+5-13=0#

#2x^2-6x-8=0" "# I happen to spot this this can be factorised

#(2x+2)(x-4) = 0#

For this to work we have #(2x+2)=0# or #(x-4)=0#

Thus we have #color(red)(x=-1" or "x=+4)# as the critical points. So These x-values are the points where #2x^2-6x+5=y=13#

So for #2x^2-6x+5>13# we have to look at the the #x's# for making #y>13# and these are: #color(green)(x<-1 and x>4) #
#color(white)("dddddddddddddddddddddddddddddddddd")color(red)(uarr)#
#color(white)("ddddddddddddddddddddd")color(red)("Corrected "x>14" to "x>4)#