10m^2-20m-48=0

Solution for 10m^2-20m-48=0 equation:

10m^2-20m-48=0

a = 10; b = -20; c = -48;
Δ = b2-4ac
Δ = -202-4·10·(-48)
Δ = 2320
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:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{145}}{2*10}=\frac{20-4\sqrt{145}}{20}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{145}}{2*10}=\frac{20+4\sqrt{145}}{20}
$