25t^2-14t-2=0

Solution for 25t^2-14t-2=0 equation:

25t^2-14t-2=0

a = 25; b = -14; c = -2;
Δ = b2-4ac
Δ = -142-4·25·(-2)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{11}}{2*25}=\frac{14-6\sqrt{11}}{50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{11}}{2*25}=\frac{14+6\sqrt{11}}{50}
$