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8y^2+10y-5=0
a = 8; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·8·(-5)
Δ = 260
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{65}}{2*8}=\frac{-10-2\sqrt{65}}{16} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{65}}{2*8}=\frac{-10+2\sqrt{65}}{16} $
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