Question 777090
There may be a typo, but as written the problem does not provide a good example to demonstrate factoring, or using the quadratic formula.
 
To solve {{{(2/3)t^2 + (-4/3)t^2 -(1/5) = 0 }}},
I would start by multiplying both sides of the equation times 15, so I do not have to deal with fractions.
{{{15*((2/3)t^2 + (-4/3)t^2 -(1/5)) = 15*0 }}}
{{{15*(2/3)t^2 + 15*(-4/3)t^2 -15*(1/5)) = 0 }}}
{{{10t^2-20t^2-3=0}}} --> {{{-10t^2-3=0}}}
Multiplying both sides by (-1), it turns into {{{10t^2+3=0}}}
Now it looks easier, but it cannot be factored. Using the quadratic formula would be more trouble than it is worth, and it is obvious that it does not have any real solution.
Real numbers, have non-negative squares, so {{{t^2>=0}}}, {{{10t^2>=0}}}, and
{{{10t^2+3>=3}}} cannot be zero.
To look for complex solutions, I would first solve for {{{t^2}}}
{{{10t^2+3=0}}}-->{{{10t^2=-3}}}-->{{{t^2=-3/10}}}
Then the solutions would be {{{system(t=i*sqrt(3/10)=i*sqrt(30)/10, and, t=-i*sqrt(3/10)=-i*sqrt(30)/10)}}}