13t=13t^2+1

Solution for 13t=13t^2+1 equation:

13t=13t^2+1

We move all terms to the left:

13t-(13t^2+1)=0

We get rid of parentheses

-13t^2+13t-1=0

a = -13; b = 13; c = -1;
Δ = b2-4ac
Δ = 132-4·(-13)·(-1)
Δ = 117
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3\sqrt{13}}{2*-13}=\frac{-13-3\sqrt{13}}{-26}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3\sqrt{13}}{2*-13}=\frac{-13+3\sqrt{13}}{-26}
$