H=-16t^2+64+60

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Solution for H=-16t^2+64+60 equation:



=-16H^2+64+60
We move all terms to the left:
-(-16H^2+64+60)=0
We get rid of parentheses
16H^2-64-60=0
We add all the numbers together, and all the variables
16H^2-124=0
a = 16; b = 0; c = -124;
Δ = b2-4ac
Δ = 02-4·16·(-124)
Δ = 7936
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:
$\sqrt{\Delta}=\sqrt{7936}=\sqrt{256*31}=\sqrt{256}*\sqrt{31}=16\sqrt{31}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{31}}{2*16}=\frac{0-16\sqrt{31}}{32} =-\frac{16\sqrt{31}}{32} =-\frac{\sqrt{31}}{2} $
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{31}}{2*16}=\frac{0+16\sqrt{31}}{32} =\frac{16\sqrt{31}}{32} =\frac{\sqrt{31}}{2} $

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