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1/3x+20=x
We move all terms to the left:
1/3x+20-(x)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
-1x+1/3x+20=0
We multiply all the terms by the denominator
-1x*3x+20*3x+1=0
Wy multiply elements
-3x^2+60x+1=0
a = -3; b = 60; c = +1;
Δ = b2-4ac
Δ = 602-4·(-3)·1
Δ = 3612
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3612}=\sqrt{4*903}=\sqrt{4}*\sqrt{903}=2\sqrt{903}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-2\sqrt{903}}{2*-3}=\frac{-60-2\sqrt{903}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+2\sqrt{903}}{2*-3}=\frac{-60+2\sqrt{903}}{-6} $
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