H(t)=-4.9t^2+30(5)+400

Solution for H(t)=-4.9t^2+30(5)+400 equation:

(H)=-4.9H^2+30(5)+400

We move all terms to the left:

(H)-(-4.9H^2+30(5)+400)=0

We get rid of parentheses

4.9H^2+H-305-400=0

We add all the numbers together, and all the variables

4.9H^2+H-705=0

a = 4.9; b = 1; c = -705;
Δ = b2-4ac
Δ = 12-4·4.9·(-705)
Δ = 13819
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{13819}=\sqrt{1*13819}=\sqrt{1}*\sqrt{13819}=1\sqrt{13819}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1\sqrt{13819}}{2*4.9}=\frac{-1-1\sqrt{13819}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1\sqrt{13819}}{2*4.9}=\frac{-1+1\sqrt{13819}}{9.8}
$