Solve the given equation below :
$\dfrac{{{t^2} + 1}}{t} = \dfrac{{58}}{{21}}$.

Hint – In order to solve this problem, make the equation given free form fractional term and solve the quadratic equation obtained by using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step-by-step answer:
The given equation is,
 $\dfrac{{{t^2} + 1}}{t} = \dfrac{{58}}{{21}}$
On solving it further we get,
$
  21{t^2} + 21 - 58t = 0 \\
  21{t^2} - 58t + 21 = 0 \\
$
The above equation is in the form,
$a{t^2} + bt + c = 0$ (i) (General equation)
And we know the roots of above equation can be written as,
$t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ (ii)
On comparing the general and obtained equation we can say,
a = 21, b = -58, c = 21 . (iii)
From (i), (ii) and (iii) we can write the roots of obtained equations by putting the values as,
$
  t = \dfrac{{ - ( - 58) \pm \sqrt {{{( - 58)}^2} - 4(21)(21)} }}{{2(21)}} \\
  t = \dfrac{{58 \pm \sqrt {3364 - 1764} }}{{42}} \\
  t = \dfrac{{58 \pm \sqrt {1600} }}{{42}} \\
$
We know square root of 1600 is 40 so we write,
$
  t = \dfrac{{58 \pm 40}}{{42}} = \dfrac{{98}}{{42}},\dfrac{{18}}{{42}} \\
  t = \dfrac{7}{3},\dfrac{3}{7} \\
$

Therefore the roots of this equation are $\dfrac{7}{3}$ and $\dfrac{3}{7}$.

Note – In these type of questions of quadratic equations in one variable first we have to simplify the given equation, such that we can compare it with the general equation and find its respective roots by applying the Sridhar Acharya’s formula($\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$) .