-4m^2+16m-1=0

Solution for -4m^2+16m-1=0 equation:

-4m^2+16m-1=0

a = -4; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·(-4)·(-1)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{15}}{2*-4}=\frac{-16-4\sqrt{15}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{15}}{2*-4}=\frac{-16+4\sqrt{15}}{-8}
$