H(t)=-16t^2+5t+227

Solution for H(t)=-16t^2+5t+227 equation:

(H)=-16H^2+5H+227

We move all terms to the left:

(H)-(-16H^2+5H+227)=0

We get rid of parentheses

16H^2-5H+H-227=0

We add all the numbers together, and all the variables

16H^2-4H-227=0

a = 16; b = -4; c = -227;
Δ = b2-4ac
Δ = -42-4·16·(-227)
Δ = 14544
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:

$\sqrt{\Delta}=\sqrt{14544}=\sqrt{144*101}=\sqrt{144}*\sqrt{101}=12\sqrt{101}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12\sqrt{101}}{2*16}=\frac{4-12\sqrt{101}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12\sqrt{101}}{2*16}=\frac{4+12\sqrt{101}}{32}
$