H=3.9t^2+15.6t

Solution for H=3.9t^2+15.6t equation:

=3.9H^2+15.6H

We move all terms to the left:

-(3.9H^2+15.6H)=0

We get rid of parentheses

-3.9H^2-15.6H=0

a = -3.9; b = -15.6; c = 0;
Δ = b2-4ac
Δ = -15.62-4·(-3.9)·0
Δ = 243.36
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{-(-15.6)-\sqrt{243.36}}{2*-3.9}=\frac{15.6-\sqrt{243.36}}{-7.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15.6)+\sqrt{243.36}}{2*-3.9}=\frac{15.6+\sqrt{243.36}}{-7.8}
$