-16t^2+26t+105=0

Solution for -16t^2+26t+105=0 equation:

-16t^2+26t+105=0

a = -16; b = 26; c = +105;
Δ = b2-4ac
Δ = 262-4·(-16)·105
Δ = 7396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7396}=86$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-86}{2*-16}=\frac{-112}{-32}
=3+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+86}{2*-16}=\frac{60}{-32}
=-1+7/8
$