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12y^2-20y+3=0
a = 12; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·12·3
Δ = 256
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{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*12}=\frac{4}{24} =1/6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*12}=\frac{36}{24} =1+1/2 $
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