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8y^2+14y+5=0
a = 8; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·8·5
Δ = 36
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{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*8}=\frac{-20}{16} =-1+1/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*8}=\frac{-8}{16} =-1/2 $
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