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15y^2-4y-4=0
a = 15; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·15·(-4)
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*15}=\frac{-12}{30} =-2/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*15}=\frac{20}{30} =2/3 $
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