12t^2-13t-4=0

Solution for 12t^2-13t-4=0 equation:

12t^2-13t-4=0

a = 12; b = -13; c = -4;
Δ = b2-4ac
Δ = -132-4·12·(-4)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-19}{2*12}=\frac{-6}{24}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+19}{2*12}=\frac{32}{24}
=1+1/3
$