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1/3x^2+3=6
We move all terms to the left:
1/3x^2+3-(6)=0
Domain of the equation: 3x^2!=0We add all the numbers together, and all the variables
x^2!=0/3
x^2!=√0
x!=0
x∈R
1/3x^2-3=0
We multiply all the terms by the denominator
-3*3x^2+1=0
Wy multiply elements
-9x^2+1=0
a = -9; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-9)·1
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-9}=\frac{-6}{-18} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-9}=\frac{6}{-18} =-1/3 $
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