H=-5t^2+26.45

Solution for H=-5t^2+26.45 equation:

=-5H^2+26.45

We move all terms to the left:

-(-5H^2+26.45)=0

We get rid of parentheses

5H^2-26.45=0

a = 5; b = 0; c = -26.45;
Δ = b2-4ac
Δ = 02-4·5·(-26.45)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-23}{2*5}=\frac{-23}{10}
=-2+3/10
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+23}{2*5}=\frac{23}{10}
=2+3/10
$