2(x^2+4x)=1

Solution for 2(x^2+4x)=1 equation:

2(x^2+4x)=1

We move all terms to the left:

2(x^2+4x)-(1)=0

We multiply parentheses

2x^2+8x-1=0

a = 2; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·2·(-1)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{2}}{2*2}=\frac{-8-6\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{2}}{2*2}=\frac{-8+6\sqrt{2}}{4}
$