4p^2-16p-48=0

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Solution for 4p^2-16p-48=0 equation:



4p^2-16p-48=0
a = 4; b = -16; c = -48;
Δ = b2-4ac
Δ = -162-4·4·(-48)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*4}=\frac{-16}{8} =-2 $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*4}=\frac{48}{8} =6 $

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