2/3h+6=3h+20

Solution for 2/3h+6=3h+20 equation:

2/3h+6=3h+20

We move all terms to the left:

2/3h+6-(3h+20)=0


Domain of the equation: 3h!=0

h!=0/3

h!=0

h∈R


We get rid of parentheses

2/3h-3h-20+6=0

We multiply all the terms by the denominator


-3h*3h-20*3h+6*3h+2=0

Wy multiply elements

-9h^2-60h+18h+2=0

We add all the numbers together, and all the variables

-9h^2-42h+2=0

a = -9; b = -42; c = +2;
Δ = b2-4ac
Δ = -422-4·(-9)·2
Δ = 1836
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:

$\sqrt{\Delta}=\sqrt{1836}=\sqrt{36*51}=\sqrt{36}*\sqrt{51}=6\sqrt{51}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{51}}{2*-9}=\frac{42-6\sqrt{51}}{-18}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{51}}{2*-9}=\frac{42+6\sqrt{51}}{-18}
$