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1/3x+5=3x-3
We move all terms to the left:
1/3x+5-(3x-3)=0
Domain of the equation: 3x!=0We get rid of parentheses
x!=0/3
x!=0
x∈R
1/3x-3x+3+5=0
We multiply all the terms by the denominator
-3x*3x+3*3x+5*3x+1=0
Wy multiply elements
-9x^2+9x+15x+1=0
We add all the numbers together, and all the variables
-9x^2+24x+1=0
a = -9; b = 24; c = +1;
Δ = b2-4ac
Δ = 242-4·(-9)·1
Δ = 612
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{17}}{2*-9}=\frac{-24-6\sqrt{17}}{-18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{17}}{2*-9}=\frac{-24+6\sqrt{17}}{-18} $
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