H=-2d^2+12d-10

Solution for H=-2d^2+12d-10 equation:

=-2H^2+12H-10

We move all terms to the left:

-(-2H^2+12H-10)=0

We get rid of parentheses

2H^2-12H+10=0

a = 2; b = -12; c = +10;
Δ = b2-4ac
Δ = -122-4·2·10
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*2}=\frac{4}{4}
=1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*2}=\frac{20}{4}
=5
$