48y^2-32y+5=0

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Solution for 48y^2-32y+5=0 equation: 



48y^2-32y+5=0

a = 48; b = -32; c = +5;
Δ = b2-4ac
Δ = -322-4·48·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8}{2*48}=\frac{24}{96}
=1/4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8}{2*48}=\frac{40}{96}
=5/12
$

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