H=16t^2+122t+25

Solution for H=16t^2+122t+25 equation:

=16H^2+122H+25

We move all terms to the left:

-(16H^2+122H+25)=0

We get rid of parentheses

-16H^2-122H-25=0

a = -16; b = -122; c = -25;
Δ = b2-4ac
Δ = -1222-4·(-16)·(-25)
Δ = 13284
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:

$\sqrt{\Delta}=\sqrt{13284}=\sqrt{324*41}=\sqrt{324}*\sqrt{41}=18\sqrt{41}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-122)-18\sqrt{41}}{2*-16}=\frac{122-18\sqrt{41}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-122)+18\sqrt{41}}{2*-16}=\frac{122+18\sqrt{41}}{-32}
$