(h-4)(h-10)=1

Solution for (h-4)(h-10)=1 equation:

(h-4)(h-10)=1

We move all terms to the left:

(h-4)(h-10)-(1)=0

We multiply parentheses ..

(+h^2-10h-4h+40)-1=0

We get rid of parentheses

h^2-10h-4h+40-1=0

We add all the numbers together, and all the variables

h^2-14h+39=0

a = 1; b = -14; c = +39;
Δ = b2-4ac
Δ = -142-4·1·39
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{10}}{2*1}=\frac{14-2\sqrt{10}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{10}}{2*1}=\frac{14+2\sqrt{10}}{2}
$