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2/3h+2=h
We move all terms to the left:
2/3h+2-(h)=0
Domain of the equation: 3h!=0We add all the numbers together, and all the variables
h!=0/3
h!=0
h∈R
-1h+2/3h+2=0
We multiply all the terms by the denominator
-1h*3h+2*3h+2=0
Wy multiply elements
-3h^2+6h+2=0
a = -3; b = 6; c = +2;
Δ = b2-4ac
Δ = 62-4·(-3)·2
Δ = 60
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*-3}=\frac{-6-2\sqrt{15}}{-6} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*-3}=\frac{-6+2\sqrt{15}}{-6} $
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