2y^2+28y+96=0

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Solution for 2y^2+28y+96=0 equation: 



2y^2+28y+96=0

a = 2; b = 28; c = +96;
Δ = b2-4ac
Δ = 282-4·2·96
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*2}=\frac{-32}{4}
=-8
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*2}=\frac{-24}{4}
=-6
$

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