2/3h=h+12

Solution for 2/3h=h+12 equation:

2/3h=h+12

We move all terms to the left:

2/3h-(h+12)=0


Domain of the equation: 3h!=0

h!=0/3

h!=0

h∈R


We get rid of parentheses

2/3h-h-12=0

We multiply all the terms by the denominator


-h*3h-12*3h+2=0

Wy multiply elements

-3h^2-36h+2=0

a = -3; b = -36; c = +2;
Δ = b2-4ac
Δ = -362-4·(-3)·2
Δ = 1320
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:

$\sqrt{\Delta}=\sqrt{1320}=\sqrt{4*330}=\sqrt{4}*\sqrt{330}=2\sqrt{330}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{330}}{2*-3}=\frac{36-2\sqrt{330}}{-6}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{330}}{2*-3}=\frac{36+2\sqrt{330}}{-6}
$