H=98t+-4.9t2

Solution for H=98t+-4.9t2 equation:

=98H+-4.9H^2

We move all terms to the left:

-(98H+-4.9H^2)=0

We use the square of the difference formula

-(98H-4.9H^2)=0

We get rid of parentheses

4.9H^2-98H=0

a = 4.9; b = -98; c = 0;
Δ = b2-4ac
Δ = -982-4·4.9·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-98}{2*4.9}=\frac{0}{9.8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+98}{2*4.9}=\frac{196}{9.8}
=20
$