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3x^2+x-2/3=0
We multiply all the terms by the denominator
3x^2*3+x*3-2=0
Wy multiply elements
9x^2+3x-2=0
a = 9; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·9·(-2)
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*9}=\frac{-12}{18} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*9}=\frac{6}{18} =1/3 $
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