2h^2+4h+1=0

Solution for 2h^2+4h+1=0 equation:

2h^2+4h+1=0

a = 2; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·2·1
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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