-2p^2=12p+16

Solution for -2p^2=12p+16 equation:

-2p^2=12p+16

We move all terms to the left:

-2p^2-(12p+16)=0

We get rid of parentheses

-2p^2-12p-16=0

a = -2; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·(-2)·(-16)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*-2}=\frac{8}{-4}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*-2}=\frac{16}{-4}
=-4
$