0=25t^2-5t-4

Solution for 0=25t^2-5t-4 equation:

0=25t^2-5t-4

We move all terms to the left:

0-(25t^2-5t-4)=0

We add all the numbers together, and all the variables

-(25t^2-5t-4)=0

We get rid of parentheses

-25t^2+5t+4=0

a = -25; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-25)·4
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{17}}{2*-25}=\frac{-5-5\sqrt{17}}{-50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{17}}{2*-25}=\frac{-5+5\sqrt{17}}{-50}
$