Question

Simplify:

$\frac{t^2+1}{t}=\frac{58}{21}$

Answer 1

Given: $\frac{t^2+1}{t}=\frac{58}{21}$

$\Rightarrow 21t^2-58t+21=0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t=\frac{58\pm \sqrt{(58{)}^2-4(21\left)\right(21)}}{2\left(21\right)}$

$t=\frac{58\pm 40}{42}$

$t=\frac{98}{42}$ or $t=\frac{18}{42}$.