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5x+3/2x+2=8
We move all terms to the left:
5x+3/2x+2-(8)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
5x+3/2x-6=0
We multiply all the terms by the denominator
5x*2x-6*2x+3=0
Wy multiply elements
10x^2-12x+3=0
a = 10; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·10·3
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:x_{1}=\frac{-b-\sqrt{\Delta}}{2a}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}
The end solution:
\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{6}}{2*10}=\frac{12-2\sqrt{6}}{20}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{6}}{2*10}=\frac{12+2\sqrt{6}}{20}
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