-16t^2+26t+13=16

Solution for -16t^2+26t+13=16 equation:

-16t^2+26t+13=16

We move all terms to the left:

-16t^2+26t+13-(16)=0

We add all the numbers together, and all the variables

-16t^2+26t-3=0

a = -16; b = 26; c = -3;
Δ = b2-4ac
Δ = 262-4·(-16)·(-3)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*-16}=\frac{-48}{-32}
=1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*-16}=\frac{-4}{-32}
=1/8
$