8y^2+18y-5=0

Solution for 8y^2+18y-5=0 equation:

8y^2+18y-5=0

a = 8; b = 18; c = -5;
Δ = b2-4ac
Δ = 182-4·8·(-5)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*8}=\frac{-40}{16}
=-2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*8}=\frac{4}{16}
=1/4
$