2x+2x^2+4=16

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

2x+2x^2+4=16

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+2x-12=0

a = 2; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·2·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*2}=\frac{8}{4}
=2
$