4p^2+16p-48=0

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