-4t^2+13t+12=0

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

-4t^2+13t+12=0

a = -4; b = 13; c = +12;
Δ = b2-4ac
Δ = 132-4·(-4)·12
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*-4}=\frac{-32}{-8}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*-4}=\frac{6}{-8}
=-3/4
$