3p^2-18p+24=0

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Solution for 3p^2-18p+24=0 equation: 



3p^2-18p+24=0

a = 3; b = -18; c = +24;
Δ = b2-4ac
Δ = -182-4·3·24
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{-(-18)-6}{2*3}=\frac{12}{6}
=2
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*3}=\frac{24}{6}
=4
$

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