-16t^2+26t+6.5=0

Solution for -16t^2+26t+6.5=0 equation:

-16t^2+26t+6.5=0

a = -16; b = 26; c = +6.5;
Δ = b2-4ac
Δ = 262-4·(-16)·6.5
Δ = 1092
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1092}=\sqrt{4*273}=\sqrt{4}*\sqrt{273}=2\sqrt{273}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{273}}{2*-16}=\frac{-26-2\sqrt{273}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{273}}{2*-16}=\frac{-26+2\sqrt{273}}{-32}
$