(2p^2)+14p+24=0

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Solution for (2p^2)+14p+24=0 equation: 



(2p^2)+14p+24=0

a = 2; b = 14; c = +24;
Δ = b2-4ac
Δ = 142-4·2·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*2}=\frac{-16}{4}
=-4
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*2}=\frac{-12}{4}
=-3
$

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