H=-16t^2+26t+

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

=-16H^2+26H+

We move all terms to the left:

-(-16H^2+26H+)=0

We get rid of parentheses

16H^2-26H-=0

We add all the numbers together, and all the variables

16H^2-26H=0

a = 16; b = -26; c = 0;
Δ = b2-4ac
Δ = -262-4·16·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-26}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+26}{2*16}=\frac{52}{32}
=1+5/8
$