H(t)=16t^2+30t+4

Solution for H(t)=16t^2+30t+4 equation:

(H)=16H^2+30H+4

We move all terms to the left:

(H)-(16H^2+30H+4)=0

We get rid of parentheses

-16H^2+H-30H-4=0

We add all the numbers together, and all the variables

-16H^2-29H-4=0

a = -16; b = -29; c = -4;
Δ = b2-4ac
Δ = -292-4·(-16)·(-4)
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-3\sqrt{65}}{2*-16}=\frac{29-3\sqrt{65}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+3\sqrt{65}}{2*-16}=\frac{29+3\sqrt{65}}{-32}
$