H=-4.9t^2+7t+40

Solution for H=-4.9t^2+7t+40 equation:

=-4.9H^2+7H+40

We move all terms to the left:

-(-4.9H^2+7H+40)=0

We get rid of parentheses

4.9H^2-7H-40=0

a = 4.9; b = -7; c = -40;
Δ = b2-4ac
Δ = -72-4·4.9·(-40)
Δ = 833
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:

$\sqrt{\Delta}=\sqrt{833}=\sqrt{49*17}=\sqrt{49}*\sqrt{17}=7\sqrt{17}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7\sqrt{17}}{2*4.9}=\frac{7-7\sqrt{17}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7\sqrt{17}}{2*4.9}=\frac{7+7\sqrt{17}}{9.8}
$