H(T)=-4.9t^2+12

Solution for H(T)=-4.9t^2+12 equation:

()=-4.9H^2+12

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

4.9H^2-12=0

a = 4.9; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·4.9·(-12)
Δ = 235.2
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{-(0)-\sqrt{235.2}}{2*4.9}=\frac{0-\sqrt{235.2}}{9.8}
=-\frac{\sqrt{}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{235.2}}{2*4.9}=\frac{0+\sqrt{235.2}}{9.8}
=\frac{\sqrt{}}{9.8}
$