16y^2-18y+5=0

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Solution for 16y^2-18y+5=0 equation:



16y^2-18y+5=0
a = 16; b = -18; c = +5;
Δ = b2-4ac
Δ = -182-4·16·5
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*16}=\frac{16}{32} =1/2 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*16}=\frac{20}{32} =5/8 $

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