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2x+16+3/2x=20
We move all terms to the left:
2x+16+3/2x-(20)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
2x+3/2x-4=0
We multiply all the terms by the denominator
2x*2x-4*2x+3=0
Wy multiply elements
4x^2-8x+3=0
a = 4; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·4·3
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*4}=\frac{4}{8} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*4}=\frac{12}{8} =1+1/2 $
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