24m^2+58m-5=0

Solution for 24m^2+58m-5=0 equation:

24m^2+58m-5=0

a = 24; b = 58; c = -5;
Δ = b2-4ac
Δ = 582-4·24·(-5)
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{-(58)-62}{2*24}=\frac{-120}{48}
=-2+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+62}{2*24}=\frac{4}{48}
=1/12
$