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8y^2-4y-12=0
a = 8; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·8·(-12)
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*8}=\frac{-16}{16} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*8}=\frac{24}{16} =1+1/2 $
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