-4p^2+16=0

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



-4p^2+16=0

a = -4; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-4)·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-4}=\frac{-16}{-8}
=+2
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-4}=\frac{16}{-8}
=-2
$

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