H=d^2+10d+5

Solution for H=d^2+10d+5 equation:

=H^2+10H+5

We move all terms to the left:

-(H^2+10H+5)=0

We get rid of parentheses

-H^2-10H-5=0

We add all the numbers together, and all the variables

-1H^2-10H-5=0

a = -1; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·(-1)·(-5)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{5}}{2*-1}=\frac{10-4\sqrt{5}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{5}}{2*-1}=\frac{10+4\sqrt{5}}{-2}
$