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1/5x+10=x
We move all terms to the left:
1/5x+10-(x)=0
Domain of the equation: 5x!=0We add all the numbers together, and all the variables
x!=0/5
x!=0
x∈R
-1x+1/5x+10=0
We multiply all the terms by the denominator
-1x*5x+10*5x+1=0
Wy multiply elements
-5x^2+50x+1=0
a = -5; b = 50; c = +1;
Δ = b2-4ac
Δ = 502-4·(-5)·1
Δ = 2520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2520}=\sqrt{36*70}=\sqrt{36}*\sqrt{70}=6\sqrt{70}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-6\sqrt{70}}{2*-5}=\frac{-50-6\sqrt{70}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+6\sqrt{70}}{2*-5}=\frac{-50+6\sqrt{70}}{-10} $
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