2p^2-16p+24=0

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Solution for 2p^2-16p+24=0 equation: 



2p^2-16p+24=0

a = 2; b = -16; c = +24;
Δ = b2-4ac
Δ = -162-4·2·24
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*2}=\frac{8}{4}
=2
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*2}=\frac{24}{4}
=6
$

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