13t^2+78t-65=0

Solution for 13t^2+78t-65=0 equation:

13t^2+78t-65=0

a = 13; b = 78; c = -65;
Δ = b2-4ac
Δ = 782-4·13·(-65)
Δ = 9464
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:

$\sqrt{\Delta}=\sqrt{9464}=\sqrt{676*14}=\sqrt{676}*\sqrt{14}=26\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(78)-26\sqrt{14}}{2*13}=\frac{-78-26\sqrt{14}}{26}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(78)+26\sqrt{14}}{2*13}=\frac{-78+26\sqrt{14}}{26}
$