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3x^2-2x+0.33=0
a = 3; b = -2; c = +0.33;
Δ = b2-4ac
Δ = -22-4·3·0.33
Δ = 0.04
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-\sqrt{0.04}}{2*3}=\frac{2-\sqrt{0.04}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{0.04}}{2*3}=\frac{2+\sqrt{0.04}}{6} $
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