4m^2+12m-1=0

Solution for 4m^2+12m-1=0 equation:

4m^2+12m-1=0

a = 4; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·4·(-1)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{10}}{2*4}=\frac{-12-4\sqrt{10}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{10}}{2*4}=\frac{-12+4\sqrt{10}}{8}
$