H=-5t2+100+1

Solution for H=-5t2+100+1 equation:

=-5H^2+100+1

We move all terms to the left:

-(-5H^2+100+1)=0

We get rid of parentheses

5H^2-100-1=0

We add all the numbers together, and all the variables

5H^2-101=0

a = 5; b = 0; c = -101;
Δ = b2-4ac
Δ = 02-4·5·(-101)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{505}}{2*5}=\frac{0-2\sqrt{505}}{10}
=-\frac{2\sqrt{505}}{10}
=-\frac{\sqrt{505}}{5}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{505}}{2*5}=\frac{0+2\sqrt{505}}{10}
=\frac{2\sqrt{505}}{10}
=\frac{\sqrt{505}}{5}
$