2x2+4x+8=22

Solution for 2x2+4x+8=22 equation:

2x^2+4x+8=22

We move all terms to the left:

2x^2+4x+8-(22)=0

We add all the numbers together, and all the variables

2x^2+4x-14=0

a = 2; b = 4; c = -14;
Δ = b2-4ac
Δ = 42-4·2·(-14)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{2}}{2*2}=\frac{-4-8\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{2}}{2*2}=\frac{-4+8\sqrt{2}}{4}
$