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1/5x+12=x
We move all terms to the left:
1/5x+12-(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+12=0
We multiply all the terms by the denominator
-1x*5x+12*5x+1=0
Wy multiply elements
-5x^2+60x+1=0
a = -5; b = 60; c = +1;
Δ = b2-4ac
Δ = 602-4·(-5)·1
Δ = 3620
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3620}=\sqrt{4*905}=\sqrt{4}*\sqrt{905}=2\sqrt{905}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-2\sqrt{905}}{2*-5}=\frac{-60-2\sqrt{905}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+2\sqrt{905}}{2*-5}=\frac{-60+2\sqrt{905}}{-10} $
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