4y^2-3+y=0

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Solution for 4y^2-3+y=0 equation: 



4y^2-3+y=0

a = 4; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·4·(-3)
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*4}=\frac{-8}{8}
=-1
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*4}=\frac{6}{8}
=3/4
$

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