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1/2f+6=3-2f
We move all terms to the left:
1/2f+6-(3-2f)=0
Domain of the equation: 2f!=0We add all the numbers together, and all the variables
f!=0/2
f!=0
f∈R
1/2f-(-2f+3)+6=0
We get rid of parentheses
1/2f+2f-3+6=0
We multiply all the terms by the denominator
2f*2f-3*2f+6*2f+1=0
Wy multiply elements
4f^2-6f+12f+1=0
We add all the numbers together, and all the variables
4f^2+6f+1=0
a = 4; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·4·1
Δ = 20
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*4}=\frac{-6-2\sqrt{5}}{8} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*4}=\frac{-6+2\sqrt{5}}{8} $
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