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12y^2-4y-1=0
a = 12; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·12·(-1)
Δ = 64
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{64}=8$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*12}=\frac{-4}{24} =-1/6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*12}=\frac{12}{24} =1/2 $
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