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8y^2+40y-2=0
a = 8; b = 40; c = -2;
Δ = b2-4ac
Δ = 402-4·8·(-2)
Δ = 1664
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{26}}{2*8}=\frac{-40-8\sqrt{26}}{16} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{26}}{2*8}=\frac{-40+8\sqrt{26}}{16} $
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