H=16t^2+138t+55

Solution for H=16t^2+138t+55 equation:

=16H^2+138H+55

We move all terms to the left:

-(16H^2+138H+55)=0

We get rid of parentheses

-16H^2-138H-55=0

a = -16; b = -138; c = -55;
Δ = b2-4ac
Δ = -1382-4·(-16)·(-55)
Δ = 15524
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:

$\sqrt{\Delta}=\sqrt{15524}=\sqrt{4*3881}=\sqrt{4}*\sqrt{3881}=2\sqrt{3881}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-138)-2\sqrt{3881}}{2*-16}=\frac{138-2\sqrt{3881}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-138)+2\sqrt{3881}}{2*-16}=\frac{138+2\sqrt{3881}}{-32}
$