H=3+t-0.25t^2

Solution for H=3+t-0.25t^2 equation:

=3+H-0.25H^2

We move all terms to the left:

-(3+H-0.25H^2)=0

We get rid of parentheses

0.25H^2-H-3=0

We add all the numbers together, and all the variables

0.25H^2-1H-3=0

a = 0.25; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·0.25·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-2}{2*0.25}=\frac{-1}{0.5}
=-2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+2}{2*0.25}=\frac{3}{0.5}
=6
$