1/3h=6;h=3

Solution for 1/3h=6;h=3 equation:

1/3h=6h=3

We move all terms to the left:

1/3h-(6h)=0


Domain of the equation: 3h!=0

h!=0/3

h!=0

h∈R


We add all the numbers together, and all the variables

-6h+1/3h=0

We multiply all the terms by the denominator


-6h*3h+1=0

Wy multiply elements

-18h^2+1=0

a = -18; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-18)·1
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*-18}=\frac{0-6\sqrt{2}}{-36}
=-\frac{6\sqrt{2}}{-36}
=-\frac{\sqrt{2}}{-6}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*-18}=\frac{0+6\sqrt{2}}{-36}
=\frac{6\sqrt{2}}{-36}
=\frac{\sqrt{2}}{-6}
$