H(d)=-4.9d^2+58d+40

Solution for H(d)=-4.9d^2+58d+40 equation:

(H)=-4.9H^2+58H+40

We move all terms to the left:

(H)-(-4.9H^2+58H+40)=0

We get rid of parentheses

4.9H^2-58H+H-40=0

We add all the numbers together, and all the variables

4.9H^2-57H-40=0

a = 4.9; b = -57; c = -40;
Δ = b2-4ac
Δ = -572-4·4.9·(-40)
Δ = 4033
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{-(-57)-\sqrt{4033}}{2*4.9}=\frac{57-\sqrt{4033}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-57)+\sqrt{4033}}{2*4.9}=\frac{57+\sqrt{4033}}{9.8}
$