H=-d2+12d+5

Solution for H=-d2+12d+5 equation:

=-H2+12H+5

We move all terms to the left:

-(-H2+12H+5)=0

We add all the numbers together, and all the variables

-(-1H^2+12H+5)=0

We get rid of parentheses

1H^2-12H-5=0

We add all the numbers together, and all the variables

H^2-12H-5=0

a = 1; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·1·(-5)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{41}}{2*1}=\frac{12-2\sqrt{41}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{41}}{2*1}=\frac{12+2\sqrt{41}}{2}
$