H(3)=16t^2+84t+100

Solution for H(3)=16t^2+84t+100 equation:

(3)=16H^2+84H+100

We move all terms to the left:

(3)-(16H^2+84H+100)=0

We get rid of parentheses

-16H^2-84H-100+3=0

We add all the numbers together, and all the variables

-16H^2-84H-97=0

a = -16; b = -84; c = -97;
Δ = b2-4ac
Δ = -842-4·(-16)·(-97)
Δ = 848
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:

$\sqrt{\Delta}=\sqrt{848}=\sqrt{16*53}=\sqrt{16}*\sqrt{53}=4\sqrt{53}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-84)-4\sqrt{53}}{2*-16}=\frac{84-4\sqrt{53}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-84)+4\sqrt{53}}{2*-16}=\frac{84+4\sqrt{53}}{-32}
$