(7y^2)+13y+4=0

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Solution for (7y^2)+13y+4=0 equation: 



(7y^2)+13y+4=0

a = 7; b = 13; c = +4;
Δ = b2-4ac
Δ = 132-4·7·4
Δ = 57
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{-(13)-\sqrt{57}}{2*7}=\frac{-13-\sqrt{57}}{14}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{57}}{2*7}=\frac{-13+\sqrt{57}}{14}
$

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