Question 834167
{{{2t^2-3t=1}}}
{{{2t^2-3t-1=0}}}
 
FACTORING:
We multiply the {{{2}}} that is in front of {{{t^2}}} times the {{{-1}}} at the end and get
{{{2*(-1)=red(-2)}}} .
Now, we look for pairs of factors of {{{red(-2)}}} that will ad up to the {{{-3}}} in front of {{{t}}}.
We cannot find any pair that works.
We find {{{2*(-1)=red(-2)}}} and {{{1*(-2)=red(-2)}}} but the sums are wrong:
{{{2+(-1)=1}}} and {{{1+(-2)=-1}}} .
If the coefficient in front of {{{t}}} was {{{1}}} or {{{-1}}} ,
we could solve {{{2t^2+t-1=0}}} or {{{2t^2-t-1=0}}} by factoring, but
we cannot solve {{{2t^2-3t-1=0}}} <--> {{{2t^2-3t=1}}} by factoring.
 
APPLYING THE QUADRATIC FORMULA:
The quadratic formula says that the solutions to {{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} .
In {{{2t^2-3t-1=0}}} we have {{{t}}} instead of {{{x}}} ,
but the name of the variable does not matter.
{{{a=2}}} is the coefficient of the term with the variable squared.
{{{b=-3}}} is the coefficient of the term with the variable not squared.
{{{c=-1}}} is the independent term.
Applying the formula with those values substituted (and {{{t}}} instead of {{{x}}} , of course) we get
{{{t = (-(-3) +- sqrt((-3)^2-4*2*(-1)))/(2*2) }}}
{{{t = (3 +- sqrt(9+8))/4 }}}
{{{highlight(t = (3 +- sqrt(17))/4)}}}
That is an ugly expression that says that the solutions are the irrational numbers
{{{(3 + sqrt(17))/4=approximately}}}{{{1.78}}}
and
{{{(3 - sqrt(17))/4=approximately}}}{{{-0.28}}} .
No wonder factoring did not work.
Factoring can only work when the solutions are rational numbers.