5f+10=(1/2)(20f-40)

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Solution for 5f+10=(1/2)(20f-40) equation:



5f+10=(1/2)(20f-40)
We move all terms to the left:
5f+10-((1/2)(20f-40))=0
Domain of the equation: 2)(20f-40))!=0
f∈R
We add all the numbers together, and all the variables
5f-((+1/2)(20f-40))+10=0
We multiply parentheses ..
-((+20f^2+1/2*-40))+5f+10=0
We multiply all the terms by the denominator
-((+20f^2+1+5f*2*-40))+10*2*-40))=0
We calculate terms in parentheses: -((+20f^2+1+5f*2*-40)), so:
(+20f^2+1+5f*2*-40)
We get rid of parentheses
20f^2+5f*2*+1-40
We add all the numbers together, and all the variables
20f^2+5f*2*-39
Wy multiply elements
20f^2+10f^2-39
We add all the numbers together, and all the variables
30f^2-39
Back to the equation:
-(30f^2-39)
We add all the numbers together, and all the variables
-(30f^2-39)=0
We get rid of parentheses
-30f^2+39=0
a = -30; b = 0; c = +39;
Δ = b2-4ac
Δ = 02-4·(-30)·39
Δ = 4680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{4680}=\sqrt{36*130}=\sqrt{36}*\sqrt{130}=6\sqrt{130}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{130}}{2*-30}=\frac{0-6\sqrt{130}}{-60} =-\frac{6\sqrt{130}}{-60} =-\frac{\sqrt{130}}{-10} $
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{130}}{2*-30}=\frac{0+6\sqrt{130}}{-60} =\frac{6\sqrt{130}}{-60} =\frac{\sqrt{130}}{-10} $

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