H=-0.5d^2+d+4

Solution for H=-0.5d^2+d+4 equation:

=-0.5H^2+H+4

We move all terms to the left:

-(-0.5H^2+H+4)=0

We get rid of parentheses

0.5H^2-H-4=0

We add all the numbers together, and all the variables

0.5H^2-1H-4=0

a = 0.5; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·0.5·(-4)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*0.5}=\frac{-2}{1}
=-2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*0.5}=\frac{4}{1}
=4
$