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2/5h+10=6h
We move all terms to the left:
2/5h+10-(6h)=0
Domain of the equation: 5h!=0We add all the numbers together, and all the variables
h!=0/5
h!=0
h∈R
-6h+2/5h+10=0
We multiply all the terms by the denominator
-6h*5h+10*5h+2=0
Wy multiply elements
-30h^2+50h+2=0
a = -30; b = 50; c = +2;
Δ = b2-4ac
Δ = 502-4·(-30)·2
Δ = 2740
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2740}=\sqrt{4*685}=\sqrt{4}*\sqrt{685}=2\sqrt{685}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{685}}{2*-30}=\frac{-50-2\sqrt{685}}{-60} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{685}}{2*-30}=\frac{-50+2\sqrt{685}}{-60} $
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