3h=3/h+40

Solution for 3h=3/h+40 equation:

3h=3/h+40

We move all terms to the left:

3h-(3/h+40)=0


Domain of the equation: h+40)!=0

h∈R


We get rid of parentheses

3h-3/h-40=0

We multiply all the terms by the denominator


3h*h-40*h-3=0

We add all the numbers together, and all the variables

-40h+3h*h-3=0

Wy multiply elements

3h^2-40h-3=0

a = 3; b = -40; c = -3;
Δ = b2-4ac
Δ = -402-4·3·(-3)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{409}}{2*3}=\frac{40-2\sqrt{409}}{6}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{409}}{2*3}=\frac{40+2\sqrt{409}}{6}
$