4m^2+2m-1=0

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Solution for 4m^2+2m-1=0 equation:



4m^2+2m-1=0
a = 4; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·4·(-1)
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*4}=\frac{-2-2\sqrt{5}}{8} $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*4}=\frac{-2+2\sqrt{5}}{8} $

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