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-7/2(2n-3)+4n=3/2(5+2n)
We move all terms to the left:
-7/2(2n-3)+4n-(3/2(5+2n))=0
Domain of the equation: 2(2n-3)!=0
n∈R
Domain of the equation: 2(5+2n))!=0We add all the numbers together, and all the variables
n∈R
-7/2(2n-3)+4n-(3/2(2n+5))=0
We add all the numbers together, and all the variables
4n-7/2(2n-3)-(3/2(2n+5))=0
We calculate fractions
4n+(-14n2/(2(2n-3)*2(2n+5)))+(-6n2/(2(2n-3)*2(2n+5)))=0
We calculate terms in parentheses: +(-14n2/(2(2n-3)*2(2n+5))), so:
-14n2/(2(2n-3)*2(2n+5))
We multiply all the terms by the denominator
-14n2
We add all the numbers together, and all the variables
-14n^2
Back to the equation:
+(-14n^2)
We calculate terms in parentheses: +(-6n2/(2(2n-3)*2(2n+5))), so:We get rid of parentheses
-6n2/(2(2n-3)*2(2n+5))
We multiply all the terms by the denominator
-6n2
We add all the numbers together, and all the variables
-6n^2
Back to the equation:
+(-6n^2)
-14n^2-6n^2+4n=0
We add all the numbers together, and all the variables
-20n^2+4n=0
a = -20; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-20)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-20}=\frac{-8}{-40} =1/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-20}=\frac{0}{-40} =0 $
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