Theresa:
This is a quadratic equation that can be solved by first putting it into the form
at^{2}+bt+c=0, and then using the quadratic formula to solve for t.
Since both the left and right side of the given equation are divisible by 5, we could simplify by multiplying both sides by 5, and then reduce the fractions to obtain
5(4t^{2})/5=5(t/5)+5(21/10) or
4t^{2}=t+21/2
We can now subtract the terms on the right from both sides bring all terms to get it into the desired form, as follows:
4t^{2}-t-21/2=0
We can now solve for t, using the quadratic formula:
t={-b + or - SQRT[b^{2}-4ac]}/2a or
t={-(-1) + or - SQRT[(-1)^{2}-(4)(4)(-21/2)]}/[(2)(4)]
t={1 + or - SQRT[1+168]}/8
t={1 + or - SQRT(169)}/8
t={1 + or - 13}/8
This can be split into two parts, as follows:
t=(1+13)/8 or t=(1-13)/8, which results in
t=14/8 or t=-12/8, or
t=7/4 or t=-3/2 (final answers)
Our work can be checked, by separately plugging the resulting values for t back into the original equation to verify that both sides of the the equation come out the same. This is left up to the student.
Let me know if you have any questions.
George T.