25t^2-20t-12=0

Solution for 25t^2-20t-12=0 equation:

25t^2-20t-12=0

a = 25; b = -20; c = -12;
Δ = b2-4ac
Δ = -202-4·25·(-12)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-40}{2*25}=\frac{-20}{50}
=-2/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+40}{2*25}=\frac{60}{50}
=1+1/5
$