13t^2-2t-7=0

Solution for 13t^2-2t-7=0 equation:

13t^2-2t-7=0

a = 13; b = -2; c = -7;
Δ = b2-4ac
Δ = -22-4·13·(-7)
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{23}}{2*13}=\frac{2-4\sqrt{23}}{26}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{23}}{2*13}=\frac{2+4\sqrt{23}}{26}
$