2m^2+4m+1=199

Solution for 2m^2+4m+1=199 equation:

2m^2+4m+1=199

We move all terms to the left:

2m^2+4m+1-(199)=0

We add all the numbers together, and all the variables

2m^2+4m-198=0

a = 2; b = 4; c = -198;
Δ = b2-4ac
Δ = 42-4·2·(-198)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-40}{2*2}=\frac{-44}{4}
=-11
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+40}{2*2}=\frac{36}{4}
=9
$