-16t^2+26t+12=0

Solution for -16t^2+26t+12=0 equation:

-16t^2+26t+12=0

a = -16; b = 26; c = +12;
Δ = b2-4ac
Δ = 262-4·(-16)·12
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-38}{2*-16}=\frac{-64}{-32}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+38}{2*-16}=\frac{12}{-32}
=-3/8
$