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6y^2-9y+3=0
a = 6; b = -9; c = +3;
Δ = b2-4ac
Δ = -92-4·6·3
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*6}=\frac{6}{12} =1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*6}=\frac{12}{12} =1 $
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