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20+1/4x=1/2x
We move all terms to the left:
20+1/4x-(1/2x)=0
Domain of the equation: 4x!=0
x!=0/4
x!=0
x∈R
Domain of the equation: 2x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/4x-(+1/2x)+20=0
We get rid of parentheses
1/4x-1/2x+20=0
We calculate fractions
2x/8x^2+(-4x)/8x^2+20=0
We multiply all the terms by the denominator
2x+(-4x)+20*8x^2=0
Wy multiply elements
160x^2+2x+(-4x)=0
We get rid of parentheses
160x^2+2x-4x=0
We add all the numbers together, and all the variables
160x^2-2x=0
a = 160; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·160·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*160}=\frac{0}{320} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*160}=\frac{4}{320} =1/80 $
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