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16y^2+6y-1=0
a = 16; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·16·(-1)
Δ = 100
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{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*16}=\frac{-16}{32} =-1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*16}=\frac{4}{32} =1/8 $
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