20m^2-2m-48=0

Solution for 20m^2-2m-48=0 equation:

20m^2-2m-48=0

a = 20; b = -2; c = -48;
Δ = b2-4ac
Δ = -22-4·20·(-48)
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-62}{2*20}=\frac{-60}{40}
=-1+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+62}{2*20}=\frac{64}{40}
=1+3/5
$