0=-16t^2+26t+4

Solution for 0=-16t^2+26t+4 equation:

0=-16t^2+26t+4

We move all terms to the left:

0-(-16t^2+26t+4)=0

We add all the numbers together, and all the variables

-(-16t^2+26t+4)=0

We get rid of parentheses

16t^2-26t-4=0

a = 16; b = -26; c = -4;
Δ = b2-4ac
Δ = -262-4·16·(-4)
Δ = 932
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:

$\sqrt{\Delta}=\sqrt{932}=\sqrt{4*233}=\sqrt{4}*\sqrt{233}=2\sqrt{233}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{233}}{2*16}=\frac{26-2\sqrt{233}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{233}}{2*16}=\frac{26+2\sqrt{233}}{32}
$