1/2h+10=h+4

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Solution for 1/2h+10=h+4 equation:



1/2h+10=h+4
We move all terms to the left:
1/2h+10-(h+4)=0
Domain of the equation: 2h!=0
h!=0/2
h!=0
h∈R
We get rid of parentheses
1/2h-h-4+10=0
We multiply all the terms by the denominator
-h*2h-4*2h+10*2h+1=0
Wy multiply elements
-2h^2-8h+20h+1=0
We add all the numbers together, and all the variables
-2h^2+12h+1=0
a = -2; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-2)·1
Δ = 152
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:
$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{38}}{2*-2}=\frac{-12-2\sqrt{38}}{-4} $
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{38}}{2*-2}=\frac{-12+2\sqrt{38}}{-4} $

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