-4m^2+20m-12=0

Solution for -4m^2+20m-12=0 equation:

-4m^2+20m-12=0

a = -4; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·(-4)·(-12)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{13}}{2*-4}=\frac{-20-4\sqrt{13}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{13}}{2*-4}=\frac{-20+4\sqrt{13}}{-8}
$