24m^2+19m=0

Solution for 24m^2+19m=0 equation:

24m^2+19m=0

a = 24; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·24·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*24}=\frac{-38}{48}
=-19/24
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*24}=\frac{0}{48}
=0
$