H(t)=1+4t-1.86t^2

Solution for H(t)=1+4t-1.86t^2 equation:

(H)=1+4H-1.86H^2

We move all terms to the left:

(H)-(1+4H-1.86H^2)=0

We get rid of parentheses

1.86H^2-4H+H-1=0

We add all the numbers together, and all the variables

1.86H^2-3H-1=0

a = 1.86; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·1.86·(-1)
Δ = 16.44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{16.44}}{2*1.86}=\frac{3-\sqrt{16.44}}{3.72}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{16.44}}{2*1.86}=\frac{3+\sqrt{16.44}}{3.72}
$