H=3+34t+16t^2

Solution for H=3+34t+16t^2 equation:

=3+34H+16H^2

We move all terms to the left:

-(3+34H+16H^2)=0

We get rid of parentheses

-16H^2-34H-3=0

a = -16; b = -34; c = -3;
Δ = b2-4ac
Δ = -342-4·(-16)·(-3)
Δ = 964
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:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{241}}{2*-16}=\frac{34-2\sqrt{241}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{241}}{2*-16}=\frac{34+2\sqrt{241}}{-32}
$