2p2+10p+8=0

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Solution for 2p2+10p+8=0 equation: 



2p^2+10p+8=0

a = 2; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·2·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*2}=\frac{-16}{4}
=-4
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*2}=\frac{-4}{4}
=-1
$

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