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16y^2-32y+7=0
a = 16; b = -32; c = +7;
Δ = b2-4ac
Δ = -322-4·16·7
Δ = 576
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{576}=24$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-24}{2*16}=\frac{8}{32} =1/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+24}{2*16}=\frac{56}{32} =1+3/4 $
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