H=16t^2+45t+4

Solution for H=16t^2+45t+4 equation:

=16H^2+45H+4

We move all terms to the left:

-(16H^2+45H+4)=0

We get rid of parentheses

-16H^2-45H-4=0

a = -16; b = -45; c = -4;
Δ = b2-4ac
Δ = -452-4·(-16)·(-4)
Δ = 1769
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{-(-45)-\sqrt{1769}}{2*-16}=\frac{45-\sqrt{1769}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1769}}{2*-16}=\frac{45+\sqrt{1769}}{-32}
$