4m(m+2)=1

Solution for 4m(m+2)=1 equation:

4m(m+2)=1

We move all terms to the left:

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

We multiply parentheses

4m^2+8m-1=0

a = 4; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·4·(-1)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{5}}{2*4}=\frac{-8-4\sqrt{5}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{5}}{2*4}=\frac{-8+4\sqrt{5}}{8}
$