H=35t-4.9t^2+12

Solution for H=35t-4.9t^2+12 equation:

=35H-4.9H^2+12

We move all terms to the left:

-(35H-4.9H^2+12)=0

We get rid of parentheses

4.9H^2-35H-12=0

a = 4.9; b = -35; c = -12;
Δ = b2-4ac
Δ = -352-4·4.9·(-12)
Δ = 1460.2
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{-(-35)-\sqrt{1460.2}}{2*4.9}=\frac{35-\sqrt{1460.2}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1460.2}}{2*4.9}=\frac{35+\sqrt{1460.2}}{9.8}
$