2y^2-16+y=0

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Solution for 2y^2-16+y=0 equation: 



2y^2-16+y=0

a = 2; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·2·(-16)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{129}}{2*2}=\frac{-1-\sqrt{129}}{4}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*2}=\frac{-1+\sqrt{129}}{4}
$

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