2x^2+8x^+4x=-16

Solution for 2x^2+8x^+4x=-16 equation:

2x^2+8x^+4x=-16

We move all terms to the left:

2x^2+8x^+4x-(-16)=0

We add all the numbers together, and all the variables

2x^2+12x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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