H(t)=-16t^2+26t+12

Solution for H(t)=-16t^2+26t+12 equation:

(H)=-16H^2+26H+12

We move all terms to the left:

(H)-(-16H^2+26H+12)=0

We get rid of parentheses

16H^2-26H+H-12=0

We add all the numbers together, and all the variables

16H^2-25H-12=0

a = 16; b = -25; c = -12;
Δ = b2-4ac
Δ = -252-4·16·(-12)
Δ = 1393
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{1393}}{2*16}=\frac{25-\sqrt{1393}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{1393}}{2*16}=\frac{25+\sqrt{1393}}{32}
$