12y^2+y-1=0

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Solution for 12y^2+y-1=0 equation: 



12y^2+y-1=0

a = 12; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·12·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*12}=\frac{-8}{24}
=-1/3
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*12}=\frac{6}{24}
=1/4
$

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