25t^2-30t-15=0

Solution for 25t^2-30t-15=0 equation:

25t^2-30t-15=0

a = 25; b = -30; c = -15;
Δ = b2-4ac
Δ = -302-4·25·(-15)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-20\sqrt{6}}{2*25}=\frac{30-20\sqrt{6}}{50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+20\sqrt{6}}{2*25}=\frac{30+20\sqrt{6}}{50}
$