H=16t^2+105t+3

Solution for H=16t^2+105t+3 equation:

=16H^2+105H+3

We move all terms to the left:

-(16H^2+105H+3)=0

We get rid of parentheses

-16H^2-105H-3=0

a = -16; b = -105; c = -3;
Δ = b2-4ac
Δ = -1052-4·(-16)·(-3)
Δ = 10833
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{-(-105)-\sqrt{10833}}{2*-16}=\frac{105-\sqrt{10833}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-105)+\sqrt{10833}}{2*-16}=\frac{105+\sqrt{10833}}{-32}
$